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1 year ago

12 tenths+ 9 tenths = tenths = ones = tenths

Mathematics

1 answer:

stiks02 [169]1 year ago

6 0

Explain: it’s equals to 20

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Simplify the expression: (3m + 1)(2)

BARSIC [14]

Answer:

6m + 2

Step-by-step explanation:

To simplify the expression you must have to use the distributive property:

(3m + 1)(2)

2 x 3m = 6m

2 x 1 = 2

6m + 2

3 0

3 years ago

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What are the x- and y-intercepts of the line?. 2x – 3y = 12. A. x-intercept –6, y-intercept 4. B. x-intercept –4, y-intercept 6.

tigry1 [53]

D. x-intercept 6, y-intercept –4.

7 0

2 years ago

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HELP AGAIN ITS MY LAST QUESTION

Lera25 [3.4K]

Answer:

$\frac{43}{5}, \frac{-7}{5}$

Step-by-step explanation:

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2 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

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

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