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8_murik_8 [283]

4 years ago

12

Can someone help me find the answer? Also tell me how u did it?

1 answer:

jok3333 [9.3K]4 years ago

4 0

Answer:

2868

Step-by-step explanation:

Well, it said that Marco's family took 12 <em>times more photos than Kevin's.</em>

So I just did 239 times 12 and I got

<u><em>2868!</em></u>

Hope this helps!

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Express the ratio as a fraction in simplest form. 15:45 2/3 3/9 1/3 5/16

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Answer:

3/9

Step-by-step explanation:

15/45

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3/9

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Which of these is the algebraic expression for "seven more than the product of three and some number?"

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3x+7 the third 1 bro

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Assume that the probability of a driver getting into an accident is 8.6%, the

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- The answer is option B. The insurance company must request a premium of $ 1095.16

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3 years ago

Which of these points lies on the line described by the equation below? y+8=-4(x+7)

timama [110]

D. (-8,-7) Simplifying<span>y + -8 = 4(x + 7) Reorder the terms: -8 + y = 4(x + 7) Reorder the terms: -8 + y = 4(7 + x) -8 + y = (7 * 4 + x * 4) -8 + y = (28 + 4x) Solving -8 + y = 28 + 4x Solving for variable 'y'. Move all terms containing y to the left, all other terms to the right. Add '8' to each side of the equation. -8 + 8 + y = 28 + 8 + 4x</span>

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4 years ago

PLEASE ILL MAKE BRAINIEST!!!

expeople1 [14]

Answer:

The area of rectangle BEFD is 180 square units.

Step-by-step explanation:

After checking the figure given, we have the following information:

$AD = 15$ , $AB = 12$ , $BC = 15$ , $CD = 12$

By Pythagorean Theorem, we determine the length of the line segment BD:

$BD = \sqrt{AB^{2}+AD^{2}}$ (1)

$BD = \sqrt{12^{2}+15^{2}}$

$BD = 3\sqrt{41}$

In addition, we know the following characteristics of the rectangle BEFD:

$BD = EF$ , $EF = EC + CF$ , $BE = FD$ (2), (3), (4)

By Pythagorean Theorem:

$BC^{2} = BE^{2}+EC^{2}$ (5)

$CD^{2} = CF^{2}+DF^{2}$ (6)

By (3), (4), (5) and (6):

$BC^{2} = BE^{2} + EC^{2}$ (7)

$CD^{2} = (EF-EC)^{2} + BE^{2}$ (8)

By (7) in (8):

$CD^{2} = (EF-EC)^{2}+ (BC^{2}-EC^{2})$

$CD^{2} = EF^{2}-2\cdot EF\cdot EC + EC^{2}+BC^{2}-EC^{2}$

$CD^{2} = EF^{2}-2\cdot EF\cdot EC +BC^{2}$

Then, we clear $EC$ :

$2\cdot EF\cdot EC = EF^{2} + BC^{2} - CD^{2}$

$EC = \frac{EF^{2}+BC^{2}-CD^{2}}{2\cdot EF}$

If we know that $EF = 3\sqrt{41}$ , $BC = 15$ and $CD = 12$ , then the length of the segment $EC$ is:

$EC = \frac{75\sqrt{41}}{41}$

And the length of the line segment $CF$ is:

$CF = EF - EC$

$CF = 3\sqrt{41}-\frac{75\sqrt{41}}{41}$

$CF = \frac{48\sqrt{41}}{41}$

And the length of the line segment $DF$ is determined by Pythagorean Theorem:

$FD = \sqrt{CD^{2}-CF^{2}}$

$FD = \sqrt{12^{2}-\left(\frac{48\sqrt{41}}{41} \right)^{2}}$

$FD = \frac{60\sqrt{41}}{41}$

And the area of the rectangle is determined by the following formula:

$A = FD\cdot EF$

$A = \left(\frac{60\sqrt{41}}{41} \right)\cdot (3\sqrt{41})$

$A = 180$

The area of rectangle BEFD is 180 square units.

4 0

3 years ago