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3 years ago

HELP PLS MARKING BRAINLIEST SHOW WORK IF YOU CAN IF NOT ITS COMPLETELY FINE JUST DO IT

Mathematics

2 answers:

Komok [63]3 years ago

6 0

Answer:

1824 in.

Step-by-step explanation:

First find the area of the two shapes you split that are the rectangle and triangle:

Rectangle Area:

48 * 32 = 1536

Area- 1536 in.

Triangle Area:

48 * 12 * 1/2

48 * 12 divided by 2

= 288 in.

Now add the two areas up:

1536 + 288 = 1824 in.

tamaranim1 [39]3 years ago

3 0

Answer:

1824

Step-by-step explanation:

I assume you want to find the area of the shape. So in order to find the area of the triangle it is height times base than divide it by 2 which looks like this

(12*48)/2 which equals 576

than to find the area of the rectangle it is height times width which is

48*32 which equals 1536

add them together and you get 1824

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Nat2105 [25]

Ithe diameter would be about 5

4 0

2 years ago

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I need help on 13, i have the answer but am not sure how to solve please help

Georgia [21]

Answer:

x³-9x²+14x+24

Step-by-step explanation:

here given : 4, -1, 6

so put negative value of each digit in the equation

(x-4)(x+1)(x-6)

(x²-4x+x-4)(x-6)

(x²-3x-4)(x-6)

(x³-3x²-4x-6x²+18x+24)

x³-9x²+14x+24

7 0

2 years ago

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Again you don’t need to answer all of them but if you can do some that would be wonderful!

Maru [420]

Answer:

17) x=7

18) x= -26

19) x= -8

20) x= 6

Step-by-step explanation:

17) is congruent

18) is supplementary

19) is congruent

20) is congruent

8 0

2 years ago

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A prism with a triangular base has a volume of 432 cubic feet. The height of the prism is 8

Rashid [163]

Answer:

9 feet

Step-by-step explanation:

Given:

A prism with a triangular base has a volume of 432 cubic feet.

The height of the prism is 8.

The triangle base has a base of 12 feet.

Question asked:

The height of the triangular base of the prism = ?

solution:

Volume of triangular prism = $Area of triangle \times height$

$432 = Area of triangle \times8$

Dividing both side by 8

$54 = Area of triangle$

$54 =$ $\frac{1}{2} \times base \times height$

$54 = \frac{1}{2} \times12\times height\\54 = \frac{12}{2} \times height$

Multiplying both side by 2

$108 = 12 \times height$

Dividing both side by 12

$9 = height$

Thus, height of the triangular base of the prism is 9

3 0

3 years ago

Given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle

Anastasy [175]

Answer:

Part 1) False

Part 2) False

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

$(x-h)^{2} +(y-k)^{2}=r^{2}$

where

(h,k) is the center and r is the radius

In this problem the distance between the center and a point on the circle is equal to the radius

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Part 1) given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x+3)^{2} +(y-4)^{2}=r^{2}$

Find the distance (radius) between the center (-3,4) and (-6,2)

substitute in the formula of distance

$r=\sqrt{(2-4)^{2}+(-6+3)^{2}}$

$r=\sqrt{(-2)^{2}+(-3)^{2}}$

$r=\sqrt{13}\ units$

The equation of the circle is equal to

$(x+3)^{2} +(y-4)^{2}=(\sqrt{13}){2}$

$(x+3)^{2} +(y-4)^{2}=13$

Verify if the point (10,4) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=10,y=4

substitute

$(10+3)^{2} +(4-4)^{2}=13$

$(13)^{2} +(0)^{2}=13$

$169=13$ -----> is not true

therefore

The point is not on the circle

The statement is false

Part 2) given the center of the circle (1,3) and a point on the circle (2,6), (11,5) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x-1)^{2} +(y-3)^{2}=r^{2}$

Find the distance (radius) between the center (1,3) and (2,6)

substitute in the formula of distance

$r=\sqrt{(6-3)^{2}+(2-1)^{2}}$

$r=\sqrt{(3)^{2}+(1)^{2}}$

$r=\sqrt{10}\ units$

The equation of the circle is equal to

$(x-1)^{2} +(y-3)^{2}=(\sqrt{10}){2}$

$(x-1)^{2} +(y-3)^{2}=10$

Verify if the point (11,5) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=11,y=5

substitute

$(11-1)^{2} +(5-3)^{2}=10$

$(10)^{2} +(2)^{2}=10$

$104=10$ -----> is not true

therefore

The point is not on the circle

The statement is false

7 0

3 years ago