Answer: The length of the rectangle is 35 in.
The width of the rectangle is 28 in.
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Explanation:
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The formula for the area, "A", of a rectangle:
Area (A) = length (L) * width (w) ;
that is: " A = L * w " ;
A = 980 in² (given);
ratio of the length to the width is: " 5 : 4 " (given);
→ Find the length (L) and the width (w).
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→ 980 in² = (5x) * (4x) ;
in which: " 980 in² " is the area of the triangle;
" 5x" = the length (L) of the rectangle, for which we shall solve;
" 4x" = the width (w) of the rectangle, for which we shall solve.
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If we find solve for "x" ; we can solve for "5x" and "4x" (the "length" and the "width", respectively); by plugging in the solved value for "x" ;
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→ 980 in² = (5x) * (4x) ;
↔ (5x) * (4x) = 980 in² ;
→ (5x) * (4x) = (5) * (4) * (x) * (x) = 20 * x² = 20x² ;
→ 20x² = 980 ;
Divide each side by "10" ; by canceling out a "0" on each side of the equation:
→ 2x² = 98 ;
Now, divide each side of the equation by "2" ;
→ 2x² / 2 = 98 / 2 ;
to get:
→ x² = 49 ;
Now, take the "positive square root" of each side of the equation; (since a "length or width" cannot be a "negative value") ;
to isolate "x" on one side of the equation; & to solve for "x" ;
→ ⁺√(x²) = √49 ;
to get:
→ x = 7 ;
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Now, we can solve for the "length" and the "width" ;
→ The length is: "5x" ;
5x = 5(7) = " 35 in " ;
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→ The width is: "4x" ;
4x = 4(7) = "28 in."
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Let us check our answer:
→ A = L * w ;
→ 980 in² = ? 35 in. * 28 in. ?? ;
Using a calculator: "35 * 28 = 980" . Yes! ;
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{ Also note: " in * in = in² " ? Yes! } .
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Answer:
C
Step-by-step explanation:
The average rate of change of f(x) in the closed interval [ a, b ] is
Here [ a, b ] = [ - 10, 10 ], thus
f(b) = f(10) = 10² + 9(10) + 18 = 100 + 90 + 18 = 208
f(a) = f(- 10) = (- 10)² + 9(- 10) + 18 = 100 - 90 + 18 = 28, thus
average rate of change = 
= 
= 9
Your answer is y=15 for your problem
I hope this helps you
(x-5)(x+1)
Answer:
Step-by-step explanation:
First let us write the given polynomial as in descending powers of x with 0 coefficients for missing items
F(x) = x^3-3x^2+0x+0
We have to divide this by x-2
Leading terms in the dividend and divisor are
x^3 and x
Hence quotient I term would be x^3/x=x^2
x-2) x^3-3x^2+0x+0(x^2
x^3-2x^2
Multiply x-2 by x square and write below the term and subtract
We get
x-2) x^3-3x^2+0x+0(x^2
x^3-2x^2
---------------
-x^2+0x
Again take the leading terms and find quotient is –x
x-2) x^3-3x^2+0x+0(x^2-x
x^3-2x^2
---------------
-x^2+0x
-x^2-2x
Subtract to get 2x +0 as remainder.
x-2) x^3-3x^2+0x+0(x^2-x-2
x^3-2x^2
---------------
-x^2+0x
-x^2+2x
-------------
-2x-0
-2x+4
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-4
Thus remainder is -4 and quotient is x^2-x-2
