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1 year ago

7

HELP ASAP 10 POINTSSSS

1 answer:

lord [1]1 year ago

6 0

Use the Pythagorean theorem (a^2 + b^2 = c^2)

7^2 + 24^2 = x^2

x^2 = 625

x = 25

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What is the y-intersept of y=4x-6

ExtremeBDS [4]

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▹ Answer

<em>y-intercept = -6</em>

▹ Step-by-Step Explanation

The format of slope is:

y = mx + b

The b represents the y-intercept which is, -6.

Hope this helps!

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7 0

3 years ago

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The value of the 5 in the ones place in the number 5,555 is how many times as much as the value of the 5 in the hundred place?

Radda [10]

100 times, because 5 times 100 equals 500

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3 years ago

Find the directional derivative of the function at the given point in the direction of the vector v. f(x, y, z) = xyz , (3, 3, 9

Anna [14]

The derivative of $f(x,y,z)$ in the direction of a vector $\mathbf u$ is

$\nabla_{\mathbf u}f=\nabla f\cdot\dfrac{\mathbf u}{\|\mathbf u\|}$

With $f(x,y,z)=xyz$ , we get

$\nabla f=(yz,xz,xy)$

and $\mathbf u=(-1,-2,2)$ ,

$\|\mathbf u\|=\sqrt{(-1)^2+(-2)^2+2^2}=3$

Then

$\nabla_{(-1,-2,2)}f(3,3,9)=(27,27,9)\cdot\dfrac{(-1,-2,2)}3=-21$

8 0

3 years ago

What are two ways to write 2a-18 in words

Rashid [163]

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6 0

2 years ago

i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = -5\sqrt 2$

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

$r^2 = x^2 + y^2$

$r = \sqrt{x^2 + y^2$

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

$r = \sqrt{(-7)^2 + (-1)^2$

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

<u>Solving the trigonometry functions</u>

$sin\ \varnothing = \frac{y}{r}$

$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

Rationalize:

$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{x}{r}$

$cos\ \varnothing = \frac{-7}{5\sqrt 2}$

Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

3 0

2 years ago