All categories

2 years ago

Solve each equation for y:

Mathematics

2 answers:

strojnjashka [21]2 years ago

4 0

Answer:

Solve each equation for y:y^2 -1=80(y-4) divided by 4=110-63 divided by y=1

Step-by-step explanation:

sammy [17]2 years ago

3 0

Step-by-step explanation:

${y}^{2} - 1 = 80$

${y}^{2} = 81$

$y = ±9$

$\frac{y - 4}{4} = 1$

$y - 4 = 4$

$y = 8$

$\frac{10 - 63}{y} = 1$

$10 - 63 = y$

$- 53 = y$

You might be interested in

Calculator

aliya0001 [1]

91.125 is the answer

5 0

3 years ago

Please answer the file question.

viktelen [127]

Answer:48 16 = ?

27 9 = 9

60 20 - 10

Step-by-step explanation:

4 0

3 years ago

WHAT DOES THIS MEAN I DONT UNDERSTAND

Art [367]

Answer:

the goal is to try and get x alone. For the 1st problem, x=3.

Step-by-step explanation:

3x + 1 = 10 Subtract 1 both sides to cancel out the 1 on the left side. 10-1 = 9

3x = 9 To get x alone here, Divide 3 on both sides. 9/3 = 3

x = 3 Now x is alone, which means x = 3.

This is right because if you plug x with 3, the sum = 10.

(3*3) + 1 =10

9 + 1 = 10

So, that's how you know 3 is the right answer.

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

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

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

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

#SPJ4

3 0

1 year ago

2.789 in expanded form

FromTheMoon [43]

It is...........

2.789=2 ones and 7 tenths 8 hundreths 9 thousandths

4 0

3 years ago