The area of shape A is 3cm2 what is the area of shape B?

Answer:

8

Step-by-step explanation:

Each side of the. decagon is equal so each side is 8

Answer:

The slope of a line passing through two points (x1, y1) and (x2, y2) is given by the formula:

m = (y2 - y1) / (x2 - x1)

Substituting the values, we get:

m = (5 - (-10)) / (6 - 3) = 15/3 = 5

Now that we have the slope, we can use the point-slope form of a linear equation to write the equation of the line:

y - y1 = m(x - x1)

Substituting the values of m, x1, and y1, we get:

y - (-10) = 5(x - 3)

Simplifying and rearranging the equation, we get:

y + 10 = 5x - 15

y = 5x - 25

Therefore, the equation of the line passing through (3,-10) and (6,5) in slope-intercept form is y = 5x - 25.

Step-by-step explanation:

#trust me bro

Answer:

If the wholesale price of the king-size mattress is $359.00, and the specialty store marks up the price by 40%, the price Debra would pay at the specialty store is:

Wholesale price + Mark-up amount = Price at specialty store

$359.00 + 40% of $359.00 = $359.00 + $143.60 = $502.60

Therefore, Debra would pay $502.60 for the mattress at the specialty store.

Answer:

A. The volume of the triangular prism can be calculated using the formula V = (1/2)bh × h, where b is the base of the triangular face and h is the height of the prism. Thus, V = (1/2)(7 cm)(10 cm) × 20 cm = 700 cubic centimeters.

B. If the height of the prism is tripled to 60 cm, then the new volume would be V' = (1/2)(7 cm)(10 cm) × 60 cm = 2100 cubic centimeters. Thus, the volume is tripled.

C. If the base of the triangular face is tripled to 21 cm, then the new volume would be V' = (1/2)(21 cm)(10 cm) × 20 cm = 2100 cubic centimeters. Thus, the volume is tripled.

D. If the height of the triangular face is tripled to 30 cm, then the new volume would be V' = (1/2)(7 cm)(30 cm) × 20 cm = 2100 cubic centimeters. Thus, the volume is tripled.

E. If all three dimensions (base, height of triangular face, and height of prism) are tripled, then the new volume would be V' = (1/2)(21 cm)(30 cm) × 60 cm = 18900 cubic centimeters. Thus, the volume is multiplied by a factor of 27.

Answer: The missing dimension of the rhombus in the given figure is Height of rhombus h  h=A/b=90/15= 6cm. so missing dimension is h=6cm

What is Dimension ?

In general, dimension refers to the measurement or size of an object, space, or quantity along a particular axis or direction. In mathematics, dimension refers to the number of coordinates needed to specify a point in a space.

What is Rhombus ?

A rhombus is a type of quadrilateral (a four-sided polygon) in which all four sides are of equal length. It is a special case of a parallelogram in which the opposite sides are parallel to each other, and its opposite angles are equal.

In the given question,

area of rhombus is A=b*h  so it can be rewritten as  h=A/b by substituting values given in question we get h= 6cm

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Using the two points you have plotted, draw the parabola. It should look like a downward-facing curve opening at the vertex (0, 7).

What is parabola?

A parabola is a symmetrical, U-shaped curve that is formed by the graph of a quadratic function.

Assuming you meant [tex]f(x) = -x^2 + 7[/tex], here's how you can graph the parabola using the parabola tool:

Find the vertex

The vertex of the parabola is located at the point (-b/2a, f(-b/2a)), where a is the coefficient of the [tex]x^2[/tex] term and b is the coefficient of the x term. In this case, a = -1 and b = 0, so the vertex is located at the point (0, 7).

Plot the vertex

Using the parabola tool, plot the vertex at the point (0, 7).

Plot a second point

To plot a second point, you can choose any x value and find the corresponding y value using the quadratic function. For example, if you choose x = 2, then [tex]f(2) = -2^2 + 7 = 3[/tex]. So the second point is located at (2, 3).

Therefore, Using the two points you have plotted, draw the parabola. It should look like a downward-facing curve opening at the vertex (0, 7).

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Complete Question:

Use the parabola tool to graph the quadratic function.

f(x) = -√² +7

Graph the parabola by first plotting its vertex and then plotting a second point on the parabola.

By answering the presented question, we may conclude that Therefore, 1  expressions cm on the map corresponds to a real distance of 3.2 km. 

In mathematics, an expression is a collection of integers, variables, and complex mathematical (such as arithmetic, subtraction, multiplication, division, multiplications, and so on) that describes a quantity or value. Phrases can be simple, such as "3 + 4," or complicated, such as They may also contain functions like "sin(x)" or "log(y)". Expressions can be evaluated by swapping the variables with their values and performing the arithmetic operations in the order specified. If x = 2, for example, the formula "3x + 5" equals 3(2) + 5 = 11. Expressions are commonly used in mathematics to describe real-world situations, construct equations, and simplify complicated mathematical topics.

Scale 1:

320000 means that 1 unit on the map represents his 320000 units in the real world.

To find the actual distance represented by 1 cm on the map, you need to convert the units to the same scale.

1 kilometer = 100000 cm

So,

1 unit on the map = 320000 units in the real world

1 cm on the map = (1/100000) km in the real world

Multiplying both sides by 1 cm gives:

1 cm on the map = (1/100000) km * 320000

A simplification of this expression:

1 cm on the map = 3.2 km

Therefore, 1 cm on the map corresponds to a real distance of 3.2 km. 

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Answer:

24 356 ÷ 5 using long division.

Step-by-step explanation:

See the image

No, the graph of tan 0, -1r/2 < 0 < 1r/2, is not asymptotic to the lines x = i and x = -i.

An asymptote is a line that a graph approaches but never crosses. The graph of tan 0, -1r/2 < 0 < 1r/2, has a period of π, meaning it repeats after every π, and will never cross the lines x = i and x = -i. This can be seen in the equation y = tan 0, where the x-values of -1r/2 and 1r/2 are replaced with the x-values of i and -i. The equation would be y = tan(i) and y = tan(-i), and the graphs of these equations would not be asymptotic to the lines x = i and x = -i.No, the graph of tan 0, -1r/2 < 0 < 1r/2, is not asymptotic to the lines x = i and x = -i.

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The common ratio (r) of this geometric sequence is found by dividing any term by its preceding term, such as:

r = a2/a1 = 6/10 = 0.6

We can use this common ratio to find any term in the sequence using the recursive formula:

a(n) = r * a(n-1)

where a(1) is the first term in the sequence, a(n) is the nth term, and a(n-1) is the (n-1)th term

Using this formula, we can find any term in the sequence. For example:

a(2) = r * a(1) = 0.6 * 10 = 6

a(3) = r * a(2) = 0.6 * 6 = 3.6

a(4) = r * a(3) = 0.6 * 3.6 = 2.16

and so on

Therefore, the complete recursive formula for this geometric sequence is:

a(n) = 0.6 * a(n-1), where a(1) = 10 and a(n) = a(n-1) for all n > 1

Answer:

10,000

Step-by-step explanation:

The first step is to divide 500 hats by 2 hours (500 ÷ 2 = 250)

Second, multiply 250 hats by 40 hours (250 x 40 = 10,000)

Answer:

$18.03

Step-by-step explanation:

10.30 x 1.7 = 18.025

round answer to 18.3

Answer: $18.03

Step-by-step explanation:

1.75 x 10.3 = 18.025

18.025 ≈ 18.03

Answer:

1.066 (3 d.p.)

Step-by-step explanation:

The volume of the solid formed by revolving a region, R, around a vertical axis, bounded by x = a and x = b, is given by:

[tex]\displaystyle 2\pi \int^b_ar(x)h(x)\;\text{d}x[/tex]

where:

[tex]\hrulefill[/tex]

We want to find the volume of the solid formed by revolving a region, R, around the y-axis, where R is bounded by:

[tex]y=\dfrac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{3}}[/tex]

[tex]y=0[/tex]

[tex]x=0[/tex]

[tex]x=1[/tex]

As the axis of rotation is the y-axis, r(x) = x.

Therefore, in this case:

[tex]r(x)=x[/tex]

[tex]h(x)=\dfrac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{3}}[/tex]

[tex]a=0[/tex]

[tex]b=1[/tex]

Set up the integral:

[tex]\displaystyle 2\pi \int^{1}_0x \cdot\dfrac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{3}}\;\text{d}x[/tex]

Take out the constant:

[tex]\displaystyle 2\pi \cdot \dfrac{1}{\sqrt{2\pi}}\int^{1}_0x \cdot e^{-\frac{x^2}{3}}\;\text{d}x[/tex]

[tex]\displaystyle \sqrt{2\pi}\int^{1}_0x \cdot e^{-\frac{x^2}{3}}\;\text{d}x[/tex]

Integrate using the method of substitution.

[tex]\textsf{Let}\;u=-\dfrac{x^2}{3}\implies \dfrac{\text{d}u}{\text{d}x}=-\dfrac{2x}{3}\implies \text{d}x=-\dfrac{3}{2x}\;\text{d}u[/tex]

[tex]\textsf{When}\;x=0 \implies u=0[/tex]

[tex]\textsf{When}\;x=1 \implies u=-\dfrac{1}{3}[/tex]

Rewrite the original integral in terms of u and du:

[tex]\displaystyle \sqrt{2\pi}\int^{-\frac{1}{3}}_0x \cdot e^{u}\cdot -\dfrac{3}{2x}\;\text{d}u[/tex]

[tex]\displaystyle \sqrt{2\pi}\int^{-\frac{1}{3}}_0 -\dfrac{3}{2}e^{u}\; \text{d}u[/tex]

[tex]-\dfrac{3\sqrt{2\pi}}{2}\displaystyle \int^{-\frac{1}{3}}_0 e^{u}\; \text{d}u[/tex]

Evaluate:

[tex]\begin{aligned}-\dfrac{3\sqrt{2\pi}}{2}\displaystyle \int^{-\frac{1}{3}}_0 e^{u}\; \text{d}u&=-\dfrac{3\sqrt{2\pi}}{2}\left[ \vphantom{\dfrac12}e^u\right]^{-\frac{1}{3}}_0\\\\&=-\dfrac{3\sqrt{2\pi}}{2}\left[ \vphantom{\dfrac12}e^{-\frac{1}{3}}-e^0\right]\\\\&=-\dfrac{3\sqrt{2\pi}}{2}\left[ \vphantom{\dfrac12}e^{-\frac{1}{3}}-1\right]\\\\&=1.06582594...\\\\&=1.066\; \sf (3\;d.p.)\end{aligned}[/tex]

Therefore, the volume of the solid is approximately 1.066 (3 d.p.).

[tex]\hrulefill[/tex]

[tex]\boxed{\begin{minipage}{3 cm}\underline{Integrating $e^x$}\\\\$\displaystyle \int e^x\:\text{d}x=e^x(+\;\text{C})$\end{minipage}}[/tex]

Answer:

3mn^4 - 6m^2n^4

The appropriate 0.5 adjusted formula for normal approximation is option (d) p(x > 8.5)

The appropriate 0.5 adjusted formula for normal approximation to approximate binomial probabilities when n is large is

P(Z > (x + 0.5 - np) / sqrt(np(1-p)))

where Z is the standard normal variable, x is the number of successes, n is the number of trials, and p is the probability of success in each trial.

To approximate binomial probability p(x > 8) when n is large, we need to use the continuity correction and find the appropriate 0.5 adjusted formula for normal approximation. Here, x = 8, n is large, and p is unknown. We first need to find the value of p.

Assuming a binomial distribution, the mean is np and the variance is np(1-p). Since n is large, we can use the following approximation

np = mean = 8, and

np(1-p) = variance = npq

8q = npq

q = 0.875

p = 1 - q = 0.125

Now, using the continuity correction, we adjust the inequality to p(x > 8) = p(x > 8.5 - 0.5)

P(Z > (8.5 - 0.5 - 8∙0.125) / sqrt(8∙0.125∙0.875))

= P(Z > 0.5 / 0.666)

= P(Z > 0.75)

Therefore, the correct option is (d) p(x > 8.5)

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The given question is incomplete, the complete question is:

To approximate binomial probability p(x > 8) when n is large, identify the appropriate 0.5 adjusted formula for normal approximation. a) p(x > 7.5) b)  p(x >= 9) c) p(x > 9) d) p(x > 8.5)

a) D. No. A person can be both male and bet on professional sports at the same time

b) If the events A and B are independent, P(A&B) = P(A) x P(B)

P(male and also bets on professional sports) = 0.484x0.13 = 0.0629

c) P(male or bets in professional sports) = P(male) + P(bets in professional sports) - P(male and also bets on professional sports)

= 0.484 + 0.13 - 0.0629

= 0.5511

d) A. The assumption was incorrect and the events are not independent

(if the were independent, the percentage would have been 6.29)

e) A. P(male or bets on professional sports = 0.484 + 0.13 - 0.081

= 0.5330

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Answer:

497/√302 = 49.7

z-score = (44-0)/49.7 = 0.88

Probability = 0.8133

Answer:

Distance in the coordinate plane iready

Step-by-step explanation:

Sure, I can help with distance in the coordinate plane!

The distance between two points (x1, y1) and (x2, y2) in the coordinate plane can be found using the distance formula:

d = √((x2 - x1)^2 + (y2 - y1)^2)

Here's an example:

Let's say we want to find the distance between the points (3, 4) and (6, 8).

We can plug these coordinates into the distance formula:

d = √((6 - 3)^2 + (8 - 4)^2)

Simplifying the expression inside the square root:

d = √(3^2 + 4^2)

d = √(9 + 16)

d = √25

d = 5

Therefore, the distance between the points (3, 4) and (6, 8) is 5 units.

The dimensions of the rectangular frame is found as : 12 and 6 inches.

When a triangle is just a right triangle, the hypotenuse square is equal to the sum of the squares of the triangle's legs.

That's a picture frame, therefore pay attention that it must be rectangular.

Hence, the triangle is really a right triangle, and the Pythagorean theorem will eventually be applied.

You are aware that the square of the hypotenuse is 20 and equals 400.

hence, a²  + b²  = 400 and...

So because length is 4 times more than the breadth, a = b + 4.

This can be resolved if "b + 4" is substituted for "a":

(b + 4)² + b²  = 400,

(b + 4)(b + 4) + b²  = 400,

b² + 8b + 16 + b²  = 400,

2b² + 8b = 384

Further solving;

b² + 4b = 192

b² + 4b - 192 = 0

(b + 16)(b - 12) = 0

Due to the fact that a length cannot be negative, b must therefore be between b - 16 or 12 (negative value not taken)

The second leg is 12 + 4 = 6.

Thus,  the dimensions of the rectangular  frame is found as : 12 and 6 inches.

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Answer:

-8: (-4, -3]

-7: (-3, -1]

(-5, -1]: (-5, -1]

(-4, -1]: (-4, -1]

[-3, 1]: [-3, 1]

(2, 1): (1, 2]

Students were asked to simplify the expression using trigonometric identities:

 A.  student A is correct; student B was confused by the division

 B.  3: cos²(θ)/(sin(θ)csc(θ)); 4: cos²(θ)

Trigonometric Identities are equality statements that hold true for all values of the variables in the equation and that use trigonometry functions.

There are several distinctive trigonometric identities that relate a triangle's side length and angle. Only the right-angle triangle is consistent with the trigonometric identities.

The six trigonometric ratios serve as the foundation for all trigonometric identities. Sine, cosine, tangent, cosecant, secant, and cotangent are some of their names.

Each student correctly made use of the trigonometric identities

cosec(θ) = 1/sin(θ)

1 -sin²(θ) = cos²(θ)

A.

Student A's work is correct.

Student B apparently got confused by the two denominators in Step 2, and incorrectly replaced them with their quotient instead of their product.

The transition from Step 2 can look like:

[tex]\frac{(\frac{1-sin^2\theta}{sin\theta} )}{cosec\theta} =\frac{1-sin^2\theta}{sin\theta} .\frac{1}{cosec\theta} =\frac{cos^2\theta}{(sin\theta)(cosec\theta)}[/tex]

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Complete question:

Students were asked to simplify the expression the quantity cosecant theta minus sine theta end quantity over cosecant period Two students' work is given. (In image below)

Part A: Which student simplified the expression incorrectly? Explain the errors that were made or the formulas that were misused. (5 points)

Part B: Complete the student's solution correctly, beginning with the location of the error. (5 points)

The true statement for all invertible matrices A and B are

1. (In−A)(In+A)=In−A².

2. (AB)⁻¹=A⁻¹B⁻¹

4. A⁷ is invertible.

The given statement is true for all invertible matrices A. To prove this statement, we can expand the left-hand side of the equation as follows:

(In−A)(In+A) = In(In) + In(A) − A(In) − A(A)

= In² + InA − AIn − A²

= In + InA − AIn − A²

= In − A²

Therefore, we have shown that (In−A)(In+A)=In−A2 is true for all invertible matrices A.

The statement is true for all invertible matrices A and B. To prove this statement, we can use the definition of the inverse of a matrix. The inverse of a matrix A is a matrix A⁻¹ such that AA⁻¹ = A⁻¹A = I, where I is the identity matrix. Using this definition, we can show that:

(AB)(A⁻¹B⁻¹) = A(BB⁻¹)A⁻¹ = AIA⁻¹ = AA⁻¹ = I

(B⁻¹A⁻¹)(AB) = B⁻¹(A⁻¹A)B = B⁻¹IB = BB⁻¹ = I

Therefore, we have shown that (AB)⁻¹ = A⁻¹B⁻¹ is true for all invertible matrices A and B.

The statement is false in general. For instance, consider the matrices A = [1 0] and B = [−1 0]. Both A and B are invertible matrices, but A + B = [0 0] which is not invertible as it is not a full rank matrix.

The statement is true for all invertible matrices A. To prove this statement, we can use the fact that the product of invertible matrices is also invertible. Since A is invertible, we can write:

A⁷ = AAAA...A

= A⁶A

= (A⁻¹)⁻¹A⁶A

= (A⁻¹A)⁻¹A⁶A

= IA⁶A

= A⁶

We can repeat this process until we get A⁷ = (A⁻¹)⁻¹. Thus, A⁷ is invertible for all invertible matrices A.

The statement is false in general. To show this, we can use a counterexample. Let A = [1 0] and B = [0 −1]. Then,

(A + B)² = [1 −1][1 −1]

= [0 0]

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Answer:

47 the answer is simply 47

Using graphs, we can see that the point (4,2) can be a coordinate where y will represent x.

The graph is simply a structured representation of the data. The numerical information gathered through observation is referred to as data.

If there is just one value of y (output) for every value of x, the relationship between x and y is said to be a function (input).

In other words, there can only be one value of y for each value of x.

Determine each plotted point's coordinates first:

(-4,4)

(-2,3)

(0,1)

(2, -1)

(3,0)

The following point cannot have any of the x-coordinates of the displayed points, which are -4, -2, 0, 2, and 3.

Options include:

A (0,1) →The relationship cannot be regarded as a function at this stage as the x-coordinate zero already has a corresponding value of y.

B (2,2) →Although there is already a value of y for the location x=2, the relationship cannot be regarded as a function at this point.

C (3,4) →Although there is already a value of y for the location x=3, the relationship cannot be regarded as a function at this point.

D (4,2) → The relationship will still be regarded as a function even though there are no points on the graph with the coordinates x=4 displayed.

Therefore, option D (4,2) is the point where y will represent x.

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The complete question:

Please see attached question

The length of JK is 18.333.

Since MN is parallel to JK, the angles formed by JLN and MLK are equal. Therefore, we can use the Triangle Proportionality Theorem, which states that if a line parallel to one side of a triangle divides the other two sides proportionally, then the triangles are similar.

Using the Triangle Proportionality Theorem, we can set up the following proportion:

[tex]$\frac{LK}{JL} = \frac{MN}{LN}$[/tex]

Therefore,

[tex]$\frac{7.2}{13.2} = \frac{10}{6}$[/tex]

We can cross-multiply to solve for JK:

[tex]$7.2 \cdot 6 = 13.2 \cdot 10$\\$43.2 = 132$\\$JK = \frac{132}{7.2} = 18.333$[/tex]

Therefore, the length of JK is 18.333.

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According to the given information, the letter that comes next in the given alphanumeric series is "N".

What is alphanumeric series?

An alphanumeric series is a sequence of letters and/or numbers that follows a certain pattern or rule. For example, "A, B, C, D, E..." is an example of an alphabetical series, and "1, 3, 5, 7, 9..." is an example of a numerical series. An alphanumeric series may combine both letters and numbers, such as "A1, B2, C3, D4, E5...". The pattern or rule followed by an alphanumeric series may be based on numerical or alphabetical order.

The given series V, Q, M, J, H, ... follows a pattern where each letter is the 6th letter from the previous letter. So, the next letter in the series would be 6 letters after H, which is N.

Therefore, the letter that comes next in the given alphanumeric series is "N".

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Answer:

A

Step-by-step explanation:

Congruent triangles

Answer:

A) IJ is congruent (equally long) to JG.

Step-by-step explanation:

KH splits IG and EF each into 2 equal halves.

the other answer options compare not-correlating distances, and so, they are not surprisingly not equally long.

Step-by-step explanation:

The formula for calculating the balance on a CD (Certificate of Deposit) after a certain amount of time is:

A = P(1 + r/n)^(nt)

Where: A = the ending balance P = the principal (initial investment) r = the annual interest rate (as a decimal) n = the number of times interest is compounded per year t = the time in years

In this case, the initial investment is $1,800.00, the annual interest rate is 2.3% (or 0.023 as a decimal), and the investment period is 2 years. Assuming that the interest is compounded annually, we can substitute these values into the formula:

A = 1800(1 + 0.023/1)^(1*2) A = 1800(1.046729) A = 1883.12

Rounding to the nearest cent, the ending balance after 2 years on the CD is $1,883.75 (option B). Therefore, option B is the correct answer.