3 years ago

(Please help) Find the area of the trapezoid.

Mathematics

1 answer:

Murljashka [212]3 years ago

5 0

Area is 5 1/4 in ^2 or 5.25 in^2
Workup in picture below
Good luck

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Round 694,553 to the nearest hundred thousand.

docker41 [41]

Answer:

pretty sure its 700,000

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3 years ago

3x-10>4+7x-3<br><br><br> Please explain.

larisa [96]

3x-10>4+7x-3

3x-10>1+7x

-4x-10>1

-4x>11

x<-11/4

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7 0

3 years ago

Read 2 more answers

I need help please don't know this

SashulF [63]

Answer:

9 is the right answer

Step-by-step explanation:

I. If c = - 4 and d = 10

then $\frac{1}{4}(-4^{3} + 10^{2} )$

= $\frac{1}{4}(-64 + 100)$

= $\frac{1}{4}(36)$

= 9

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3 years ago

Hi guys can you please help me with this test.

Anika [276]

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3 years ago

When the sum of \, 528 \, and three times a positive number is subtracted from the square of the number, the result is \, 120. F

aleksandr82 [10.1K]

Let $x$ be the unknown number. So, three times that number means $3x$ , and the square of the number is $x^2$

We have to sum 528 and three times the number, so we have $528+3x$

Then, we have to subtract this number from $x^2$ , so we have

$x^2-(3x+528)$

The result is 120, so the equation is

$x^2 - 3x - 528 = 120 \iff x^2 - 3x - 648 = 0$

This is a quadratic equation, i.e. an equation like $ax^2+bx+c=0$ . These equation can be solved - assuming they have a solution - with the following formula

$x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

If you plug the values from your equation, you have

$x_{1,2} = \dfrac{3\pm\sqrt{9-4\cdot(-648)}}{2} = \dfrac{3\pm\sqrt{9+2592}}{2} = \dfrac{3\pm\sqrt{2601}}{2} = \dfrac{3\pm51}{2}$

So, the two solutions would be

$x = \dfrac{3+51}{2} = \dfrac{54}{2} = 27$

$x = \dfrac{3-51}{2} = \dfrac{-48}{2} = -24$

But we know that x is positive, so we only accept the solution $x = 27$

6 0

3 years ago