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3 years ago

Find the equation of a line having slope -8 and y intercept -3

Mathematics

1 answer:

Andre45 [30]3 years ago

5 0

Answer:

y = -8x -3

Step-by-step explanation:

To find the y intercept of a basic y = mx + b equation you simply find the constant (In Algebra, a constant is a number on its own, or sometimes a letter such as a, b or c to stand for a fixed number). To find the slope of a basic y = mx + b equation find the value of m. The value of m will be the slope.

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Find one positive and one negative angle coterminal with an angle of 166 Question 4 options:

vesna_86 [32]

Answer: Choice A) 526 degrees, -194 degrees

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Work Shown:

A coterminal angle points in the same exact direction as the original angle.

Because there are 360 degrees in a circle, this means we can add 360 to the original angle to get 166+360 = 526, which is one positive coterminal angle to 166 degrees.

Subtract 360 from the original angle and we'll get a negative coterminal angle

166 - 360 = -194

8 0

3 years ago

With the introduction of decimals, Max is very confused about the role zeros play in a number. For example 3 and 30 are not equa

olga2289 [7]

.3 is represented as a tenth it’s 3 out of ten, but .03 is a hundredth representing as a 3 out of a hundred it’s a relatively smaller number.

7 0

3 years ago

Plz help I have no clue how to do 1 and 2

Elan Coil [88]

1) y= 1/2x+2 and I have no idea about 2

3 0

3 years ago

Find the length of UC on the attached diagram. Thank you!

ValentinkaMS [17]

<h2>Answer </h2>

The length of UC is 18

<h2>Explanation </h2>

First we are going to find the length of JN; then we are subtracting from it the length of JU plus the length of CN.

We can infer from our picture that JN is 82 + 105, so JN = 187

We can also infer that JU = JH + HU

JU = 22 + 96

JU = 118

We can also infer that CN = 51

Now we can fin the length of UC:

$UC=JN-(JU+CN)$

$UC=187-(118+51)$

$UC=18$

We can conclude that the length of UC is 18.

3 0

3 years ago

Read 2 more answers

compute the projection of → a onto → b and the vector component of → a orthogonal to → b . give exact answers.

Nina [5.8K]

$\text { Saclar projection } \frac{1}{\sqrt{3}} \text { and Vector projection } \frac{1}{3}(\hat{i}+\hat{j}+\hat{k})$

We have been given two vectors $\vec{a}$ and $\vec{b}$ , we are to find out the scalar and vector projection of $\vec{b}$ onto $\vec{a}$

we have $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$

The scalar projection of $\vec{b}$ onto $\vec{a}$ means the magnitude of the resolved component of $\vec{b}$ the direction of $\vec{a}$ and is given by

The scalar projection of $\vec{b}$ onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}$

$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\sqrt{1^2+1^1+1^2}} \\&=\frac{1^2-1^2+1^2}{\sqrt{3}}=\frac{1}{\sqrt{3}}\end{aligned}$

The Vector projection of $\vec{b}$ onto $\vec{a}$ means the resolved component of $\vec{b}$ in the direction of $\vec{a}$ and is given by

The vector projection of $\vec{b}$ onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \cdot(\hat{i}+\hat{j}+\hat{k})$

$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\left(\sqrt{1^2+1^1+1^2}\right)^2} \cdot(\hat{i}+\hat{j}+\hat{k}) \\&=\frac{1^2-1^2+1^2}{3} \cdot(\hat{i}+\hat{j}+\hat{k})=\frac{1}{3}(\hat{i}+\hat{j}+\hat{k})\end{aligned}$

To learn more about scalar and vector projection visit:brainly.com/question/21925479

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3 0

1 year ago