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2 years ago

Can someone help me with this please? I don't know how to do it, and I need the answer asap ;-; (Image Below)

Mathematics

2 answers:

Ostrovityanka [42]2 years ago

7 0

There isn’t a picture

Marizza181 [45]2 years ago

4 0

Answer:

ma'am or sir... there is no picture attached... but whats the question?

Step-by-step explanation:

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A trapezoid has a height of 6 inches, an area of 120 square inches, and one base measuring 13 inches. Find the length of the oth

kirill [66]

Answer:

Step-by-step explanation:

Givens

b1 = 13

b2 = ?

h = 6

Area = 120

Formula

Area = (b1 + b2) * h/2 Multiply by 2

2Area = (b1 + b2)*h Divide by h

2Area/h = b1 + b2 Subtract b1 from both sides

2Area/h - b1 = b2

Solution

2*120 / 6 - 13 = b2

40 - 13 = b2

b2 = 27

It is always handy to solve an equation in the form that finds the unknown on one side. It makes the solution much easier.

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2 years ago

Plz help this is a true or false correct answer only question also giving the points it's worth

Bond [772]

Answer:

1.False

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4 0

2 years ago

Read 2 more answers

X-4y=28 2x+y=2 Elimination

makvit [3.9K]

$\left\{\begin{array}{ccc}x-4y=28\\2x+y=2&\text{multiply both sides by 4}\end{array}\right\\\\\underline{+\left\{\begin{array}{ccc}x-4y=28\\8x+4y=8\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad9x=36\qquad\text{divide both sides by 9}\\.\qquad\boxed{x=4}\\\\\text{Put the value of x to the second equation:}\\\\2(4)+y=2\\8+y=2\qquad\text{subtract 8 from both sides}\\\boxed{y=-6}\\\\Answer:\ \boxed{x=4\ and\ y=-6\to(4,\ -6)}$

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2 years ago

Please help me with this homework

Sergio039 [100]

39) y=20x+175

40) y=20(2)+175, so $215

5 0

2 years ago

Why the derivative of (x^2/a^2) = (2x/a^2)?

Neko [114]

I assume you're referring to a function,

$f(x) = \dfrac{x^2}{a^2}$

where <em>a</em> is some unknown constant. By definition of the derivative,

$\displaystyle f'(x) = \lim_{h\to0}{f(x+h)-f(x)}h$

Then

$\displaystyle f'(x) = \lim_{h\to0}{\frac{(x+h)^2}{a^2}-\frac{x^2}{a^2}}h \\\\ f'(x) = \frac1{a^2} \lim_{h\to0}{(x+h)^2-x^2}h \\\\ f'(x) = \frac1{a^2} \lim_{h\to0}{(x^2+2xh+h^2)-x^2}h \\\\ f'(x) = \frac1{a^2} \lim_{h\to0}{2xh+h^2}h \\\\ f'(x) = \frac1{a^2} \lim_{h\to0}(2x+h) = \boxed{\frac{2x}{a^2}}$

8 0

2 years ago