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2 years ago

9

Solve for x. 512≥1.6x x≥3716 x≤3716 x≥845 x≤845

1 answer:

GalinKa [24]2 years ago

6 0

I'm pretty sure the answer will just be the 1.6 multiply by 512

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A school fair ticket costs $8 per adult and $1 per child. On a certain day, the total number of adults (a) and children (c) who

Jobisdone [24]

The right equations are a+c=30 (total number of people) and 8a+c=100 (total amount of money). So answers B and D cannot be right.

a+c=30 can be rewritten as c=30-a and filled into the second one:

8a + 30-a = 100 => 7a = 70 => a=10

That leaves answer A as the right one.

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3 years ago

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Answer:

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In an isosceles trapezoid the length of a diagonal is 25 cm and the length of an altitude is 15 cm. Find the area of the trapezo

andreev551 [17]

Answer:

The area of the trapezoid is $525\ cm^{2}$

Step-by-step explanation:

we know that

The area of a isosceles trapezoid is equal to the area of two isosceles right triangles plus the area of a rectangle

step 1

<em>Find the area of the isosceles right triangle</em>

Remember that

In a isosceles right triangle the height is equal to the base of the triangle

we have

$h=15\ cm$

so

$b=15\ cm$

The area is equal to

$A=\frac{1}{2}(b)(h)$

substitute the values

$A=\frac{1}{2}(15)(15)=112.5\ cm^{2}$

step 2

Find the area of the rectangle

The area of the rectangle is equal to

$A=LW$

we have

$W=15\ cm$ -----> is the height of the trapezoid

$d=25\ cm$ -----> the diagonal of the rectangle

Applying the Pythagoras Theorem

$25^{2}=L^{2}+15^{2}\\L^{2}=25^{2}-15^{2} \\ L^{2} =400\\L=20\ cm$

The area of the rectangle is

$A=(20)(15)=300\ cm^{2}$

step 3

Find the area of the trapezoid

$A=2(112.5\ cm^{2})+300\ cm^{2}=525\ cm^{2}$

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4 years ago

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Alexxx [7]

See attached for the solution

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3 years ago

A store is selling a backpack that is $18. Tax is 6%. What is the total price, including tax?

Masteriza [31]

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