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3 years ago

What is the answer too 11 - 18s = - 15?

Mathematics

1 answer:

Arlecino [84]3 years ago

4 0

Answer: s = 1.44

Step-by-step explanation:

11 - 18*s = -15

-18*s = -15 - 11

-18*s = -26

18*s = 26

s = 26/18

s = 13/9

s = 1.44

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To avoid a storm, a passenger-jet pilot descended 0.44 mile in 0.8 minutes. What was the planes average change of altitude per m

Tcecarenko [31]

The average change of altitude per minute would simply be the ratio of distance descended over time. That is:

average change of altitude = distance / time

average change of altitude = 0.44 mile / 0.8 minutes

<span>average change of altitude = 0.55 mile / minute</span>

6 0

2 years ago

2. A sphere has surface area 16π. Find the radius of the sphere.

KiRa [710]

Answer:

The answer to your question is: 2.29 units

Step-by-step explanation:

Data

Volume = 16π

radius = ?

Formula

V = $\frac{4}{3}π r^{3}$

Process

16 π = $\frac{4}{3}π r^{3}$

$\frac{16(3)}{4} = r^{3}$

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r = 2.29

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2 years ago

A protective cover for a wireless tablet has a height of 8 inches and base of 7 inches. What is the area of the cover?

RUDIKE [14]

So to find the answer of the area, you have to multiply 7 and 8 which would equal to 56.

6 0

2 years ago

Read 2 more answers

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:

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