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kari74 [83]
2 years ago
5

Bill is 6 inches taller than Nancy. Use n for Nancy's height. Write an expression for Bill's height: I​

Mathematics
1 answer:
Mazyrski [523]2 years ago
6 0

n + 6

(Please mark me brainliest! thank you in advance!!)

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I think it is 395.20 / 5 = 79.04, rounded down to 79
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My friend Coraz bought some hopts for $6 each and some rews for $4 each. The total cost was $120. Write an equation in standard
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7,8

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A farmer had a total of 995 horses and cows. After selling some of the horses and cows, he had 510 horses and 390 cows left. How
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A dinner plate has a circumference of 106.76 cm. What is the area of the dinner plate? Use = 3.14.
Sphinxa [80]
C=2pr,  r=c/(2p)

a=pr^2, using r found above we get:

a=p(c^2/(4p^2))

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a=(106.76^2)/(4*3.14)

a=11612.2176/12.56 cm^2

a≈924.54 cm^2  (to nearest one-hundredth of a square cm)
3 0
2 years ago
Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$
Anna [14]

Answer:

\large\boxed{1\dfrac{1}{3}\ u^2}

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}

6 0
3 years ago
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