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3 years ago

-15 + w = 6 can anyone help me out

Mathematics

2 answers:

Nat2105 [25]3 years ago

5 0

The answer should be w=21

pychu [463]3 years ago

3 0

Answer:

w=21

Step-by-step explanation:

w=6+15

=w=21

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How many solutions does the equation 3(x-5)-7=-3x8 have

LenaWriter [7]

Answer:

not really good at maths. 15.33

Step-by-step explanation:

3(x-5)-7=3×8

3x-15-7=24

3x =24+15+7

3x = 46

Three will divide it's self to give you one and divide 46 to give you 15.333

(correct me if i'm wrong)

8 0

3 years ago

An item that was selling 72 is reduced to 60. Find the recent decrease in price. Round you answer to the nearest tenth.

Lubov Fominskaja [6]

You need to find what percentage the difference is of 72.

First, subtract 60 from 72:

72-60=8

Next, divide 8 by 72:

8/72=0.1111

To convert this decimal to a percentage, multiply it by 100:

100*0.1111=11.11

The percent decrease is 11.11%.

Hope this helps!!

8 0

3 years ago

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

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

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