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3 years ago

A train travels at a constant rate of 30 miles per hour. how long will it take the train to travel 100 miles

Mathematics

2 answers:

Leno4ka [110]3 years ago

8 0

Answer:

it'll maybe take 3 in a half hours or so i don't know if i'm right .

Step-by-step explanation:

o-na [289]3 years ago

7 0

Answer:

3 hours and 20 min

Step-by-step explanation:

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Can anyone help me with this?0

sergeinik [125]

Answer:

C) (y - 1)(y - 5)

Step-by-step explanation:

Hi there!

First you separate your expression into groups.

$(y^2-5y)(-1y+5)$

Now we take the common multiple out of each.

$y(y - 5)-1(y - 5)$

Since we have two (y - 5)s, we can pull it out of the equation as its own common multiple.

After we do that, we are left with y - 1. (we got this from taking the common multiples out)

Now we are just left with (y - 5)(y - 1)

Hope this helped!

4 0

2 years ago

What is the solution of the system of equations shown in the graph?

Rudiy27

Answer:

The solution of the system of equations shown in the graph is (2,0)

Step-by-step explanation:

It's easy

You just look for the point of intersections the lines make.

The point is on the number 2 on the x axis, and number 0 on the y axis. Therefore its (2,0).

Happy to help!

3 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

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

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