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BaLLatris [955]
3 years ago
7

Simplified or solved for x ????

Mathematics
1 answer:
Leya [2.2K]3 years ago
5 0
\dfrac{16 \pm  \sqrt{-32} }{2}

\dfrac{16 \pm  \sqrt{-16 \times 2} }{2}

\dfrac{16 \pm 4i\sqrt{2} }{2}

8 \pm 2i \sqrt{2}
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State 4 solutions for the equation that is modeled by the graph.
Norma-Jean [14]

Answer:

see explanation

Step-by-step explanation:

Any coordinate point that lies on the line is a solution , that is

(- 2, 8 ) , (0, 4 ) , (2, 0 ) , (4, - 4 )

7 0
2 years ago
C.) 4. Solve the system of equations
raketka [301]
You can set them equal to each other so -3x+4=4x-10 and then you add 3x and 10 on both sides and get7x=14 and then divided both sides by 7 and get x = 2 and check by plugging in and you get -2 for y on both so solution is x=2
7 0
2 years ago
What is suface area of a cone with 16in radius and 38in hight?
tresset_1 [31]
2876.74665 in


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7 0
3 years ago
Salma received a $70 gift card for a coffee store. She used it in buying some coffee that cost $7.61 per pound. After buying the
rosijanka [135]

Answer:

4

Step-by-step explanation:

Amount spent on coffee = 70 - 39.56 = 30.44

One pound costs 7.61, how many will cost 30.44

7.61 : 1 = 30.44 : X

7.61 = 30.44/X

X = 30.44/7.61

X = 4

7 0
2 years ago
Read 2 more answers
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
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