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notsponge [240]
3 years ago
15

the length of a rectangle is 7 inches greater than three times its width write an expression to represent its area

Mathematics
1 answer:
icang [17]3 years ago
3 0

Step-by-step explanation:

Let the length be L.

Let the breath be B.

So, we are given that

length of a rectangle is 7 inches greater than three times its width .

So, according to the Question,

L = 7+ 3B.

So, we know that:-

Area = Length× Breadth.

So, it will be

B × (7 + 3B) = Area.

So,

3B² + 7 is an expression to represent its area.

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Lostsunrise [7]

Answer:

x=2,y=0

(You did not provide enough information for me to know what to do with said equations, so I'm assuming it was System of Equations.)

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2 years ago
Can someone please help me with this
lubasha [3.4K]
Baso 6 divided by 2 is 3 that’s what the question is saying
8 0
3 years ago
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Help ASAP can u explain a little too? (pls help if u don’t know it don’t answer pls) Thank You:)
larisa [96]
Since they are similar, the dimensions are in the same ratio. L1 = 5, L2 = 15, so they are in a 3:1 ratio. So if V1 = 60, then W1×H1 = 60/5 = 12
W2 must also be 3×W1 and H2 3×H1, and
3×3 = 9. So take 12×9 (W×H1×9) ×15 (L2) = V2
V2 = 12×9×15 = 1620 cm^3

Let me know the right answer when you find out!
8 0
3 years ago
Calculate the total area of the shaded region.
LUCKY_DIMON [66]

so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

\stackrel{f(x)}{2x^3-x^2-5x}~~ = ~~\stackrel{g(x)}{-x^2+3x}\implies 2x^3-5x=3x\implies 2x^3-8x=0 \\\\\\ 2x(x^2-4)=0\implies x^2=4\implies x=\pm\sqrt{4}\implies x= \begin{cases} 0\\ \pm 2 \end{cases}

so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill

\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}

7 0
2 years ago
What is the area of this trapezoid?
sasho [114]

Area of trapezoid : (Base 1 + base 2) x h x 1/2

Let's solve!

( 2+6) x 5 X1/2

= 40x1/2

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20 in squared is the area!

8 0
3 years ago
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