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ValentinkaMS [17]
3 years ago
15

Lil fun for yall what is the answer to this step by step?

Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
7 0

Answer:

194332.125 is the amount of interest

Step-by-step explanation:

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From 56 people to 63 people
kramer

Answer:

uhh 7 people joined?

Step-by-step explanation:

8 0
2 years ago
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Y+5<br> ——=0 solve for y<br> X-1
Ira Lisetskai [31]

Answer:

y = -5

Step-by-step explanation:

see the pic for the steps . .

5 0
4 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

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

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

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

#SPJ4

3 0
1 year ago
Rewrite f(x) = 2(x − 1)2 + 3 from vertex form to standard form. (2 points)
Ostrovityanka [42]

Answer:

F(x)= 4x-1 I think so hope it will help you

7 0
3 years ago
A 2.2 kg ball strikes a wall with a velocity of 7.4 m/s to the left. The ball bounces off with a velocity of 6.2 m/s to the righ
Naya [18.7K]

Answer:

The constant force exerted on the ball by the wall is 119.68 N.

Step-by-step explanation:

Consider the provided information.

It is given that the mass of the ball is m = 2.2 kg

The initial velocity of the ball towards left is 7.4 m/s

So the momentum of the ball when it strikes is = 2.2\times 7.4=16.28

The final velocity of the ball is -6.2 m/s

So the momentum of the ball when it strikes back is = 2.2\times -6.2=-13.64

Thus change in moment is: 16.28-(-13.64)=29.92

The duration of force exerted on the ball t = 0.25 s

Therefore, the constant force exerted on the ball by the wall is:

\frac{29.92}{0.25}=119.68

Hence, the constant force exerted on the ball by the wall is 119.68 N.

6 0
3 years ago
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