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3 years ago

I need to know the answer for my test pls help

Mathematics

2 answers:

EleoNora [17]3 years ago

7 0

Answer:

That is an existing variable such that it means the weight room has an existing are of 480 ft^2

DochEvi [55]3 years ago

5 0

3= -1 + 7y x 10 to the second power

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Which are true of the function f(x)=49(1/7)^x

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All you need to do is multiply

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3 years ago

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Center (8,9) and radius r=10

LekaFEV [45]

Answer:

A. Standard form: $(x-8)^2+(y-9)^2=100$

B. General form: $x^2+y^2-16x-18y+45=0$

Step-by-step explanation:

We have been given that center of a circle is at point (8,9) and radius of our circle is 10 units. We are asked to write the equation of our circle.

A. Since we know that the equation of a circle in standard form is: $(x-h)^2+(y-k)^2=r^2$ , where,

(x,y) = Any point on circle,

(h,k) = Center of the circle,

r = Radius of the circle.

Upon substituting our given values in standard form of circle's equation we will get,

$(x-8)^2+(y-9)^2=10^2$

$(x-8)^2+(y-9)^2=100$

Therefore, the equation of our given circle in standard form will be $(x-8)^2+(y-9)^2=100$ .

B. Since we know that equation of a circle in general form is: $x^2+y^2+Ax+By+C=0$ , where, A, B and C are constants.

Upon expanding our standard form of equation we will get,

$x^2-16x+64+y^2-18y+81=100$

$x^2-16x+64+y^2-18y+81-100=100-100$

$x^2-16x+y^2-18y+64+81-100=0$

$x^2-16x+y^2-18y+45=0$

$x^2+y^2-16x-18y+45=0$

Therefore, the equation of our given circle in general form will be $x^2+y^2-16x-18y+45=0$ .

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3 years ago

Solve abc given that C=32 b=8 a=3

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The answer to (a)(b)(c) is 2,656

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4 years ago

Circle O and circle P are congruent. Given that arc AB is congruent to arc DE, what can you conclude from the diagram?

wlad13 [49]

Answer:

Central angle is equal in both circles

And the two triangles inscribed in circle is congruent.

Step-by-step explanation:

We have been given the two circles which are congruent O and P.

Ab is congruent to DE

So, the central angle of two circles will be equal circles being congruent.

And the two triangles AOB and EPD are congruent

Radius of the congruent circle is equal

Hence, OB=PD

AB=DE

And ∠BOA=∠DPE

By SAS property ΔAOB ≅ΔEPD

5 0

4 years ago

Pyramid A has a triangular base where each side measures 4 units and a volume of 36 cubic units. Pyramid B has the same height,

omeli [17]

Answer:

The volume of pyramid B is 81 cubic units

Step-by-step explanation:

Given

Pyramid A

$s = 4$ -- base sides

$V = 36$ -- Volume

Pyramid B

$s = 6$ --- base sides

Required

Determine the volume of pyramid B [Missing from the question]

From the question, we understand that both pyramids are equilateral triangular pyramids.

The volume is calculated as:

$V = \frac{1}{3} * B * h$

Where B represents the area of the base equilateral triangle, and it is calculated as:

$B = \frac{1}{2} * s^2 * sin(60)$

Where s represents the side lengths

First, we calculate the height of pyramid A

For Pyramid A, the base area is:

$B = \frac{1}{2} * s^2 * sin(60)$

$B = \frac{1}{2} * 4^2 * \frac{\sqrt 3}{2}$

$B = \frac{1}{2} * 16 * \frac{\sqrt 3}{2}$

$B = 4\sqrt 3$

The height is calculated from:

$V = \frac{1}{3} * B * h$

This gives:

$36 = \frac{1}{3} * 4\sqrt 3 * h$

Make h the subject

$h = \frac{3 * 36}{4\sqrt 3}$

$h = \frac{3 * 9}{\sqrt 3}$

$h = \frac{27}{\sqrt 3}$

To calculate the volume of pyramid B, we make use of:

$V = \frac{1}{3} * B * h$

Since the heights of both pyramids are the same, we can make use of:

$h = \frac{27}{\sqrt 3}$

The base area B, is then calculated as:

$B = \frac{1}{2} * s^2 * sin(60)$

Where

$s = 6$

So:

$B = \frac{1}{2} * 6^2 * sin(60)$

$B = \frac{1}{2} * 36 * \frac{\sqrt 3}{2}$

$B = 9\sqrt 3$

So:

$V = \frac{1}{3} * B * h$

Where

$B = 9\sqrt 3$ and $h = \frac{27}{\sqrt 3}$

$V = \frac{1}{3} * 9\sqrt 3 * \frac{27}{\sqrt 3}$

$V = \frac{1}{3} * 9 * 27$

$V = 81$

6 0

3 years ago

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