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2 years ago

I need help with number 46 Thankyou

Mathematics

1 answer:

Over [174]2 years ago

8 0

A²+B²=C²
6²+8²=C²
C²=100
C=10
Therefore the length of AB is 10 units

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Find the geometric mean between 18 and 7. Write your answer in radical form.

solmaris [256]

Answer: Choice A

Work Shown:

$m = \text{geometric mean of x and y}\\\\m = \sqrt{x*y}\\\\m = \sqrt{18*7}\\\\m = \sqrt{9*2*7}\\\\m = \sqrt{9}*\sqrt{2*7}\\\\m = 3\sqrt{14}\\\\$

3 0

2 years ago

KA AGAIN ( 2 pdfs) correct answers only

Reil [10]

Answer:

Step-by-step explanation:

To find the constant of proportionality between y and x, divide $\frac{y}{x}$ for all given values of the chart. If you get the same number, then that is the constant. If they are all different numbers, then one does not exist for that function.

Using the method above, you should get:

Chart A = 2/5

Chart B = 1/5 (Correct Answer!)

Chart C = N/A

Hope this helps!

4 0

2 years ago

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G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =

Svetach [21]

Presumably you should be doing this using calculus methods, namely computing the surface integral along $\mathbf r(u,v)$ .

But since $\mathbf r(u,v)$ describes a sphere, we can simply recall that the surface area of a sphere of radius $a$ is $4\pi a^2$ .

In calculus terms, we would first find an expression for the surface element, which is given by

$\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k$
$\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j$
$\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k$
$\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u$

So the area of the surface is

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u$
$=-2\pi a^2(\cos\pi-\cos 0)$
$=-2\pi a^2(-1-1)$
$=4\pi a^2$

as expected.

6 0

2 years ago

Below are the data collected from two random samples of 500 American students on the number of hours they spend in school per da

balu736 [363]

Answer: Tara thinks correctly.

Step-by-step explanation:

Let's calculate the average value in sample A:

(4 × 70 + 5 × 100 + 6 × 125 + 7 × 135 + 8 × 70) ÷ 500 = 6,07 ≈ 6 hours

Let's calculate the average value in sample B:

(4 × 80 + 5 × 90 + 6 × 120 + 7 × 125 + 8 × 85) ÷ 500 = 6,09 ≈ 6 hours

So, Tara is correct in saying that the mean is 6.

3 0

3 years ago

Carlos finds the absolute value of -5.3, and then finds the opposite of his answer. Jason finds the opposite of -5.3 and then fi

polet [3.4K]

The absolute value is how far away a number is from the number line, so it is always positive. If Carlos finds the opposite of the absolute value, that means he found the negative version. So Jason's final value will be greater.

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2 years ago

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