2ci 1=-d 6-ci solve for c and d

1 answer:

5 0

Imaginary numbers ? just equate the imaginary with imaginary and real with real

<span>2ci 1=-d 6-ci <<< what the * is that ?

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Evaluate the following expression using the values given: Find −a2 − 3b3 + c2 + 2b3 − c2 if a = 3, b = 2, and c = −3. (1 point)

wariber [46]

The answer is -12
because you just have to simplyfy it and plug in the values

6 0

3 years ago

A 5-pound bag of apples cost $3.50. Use that rate, to complete the table to identify equivalent ratios, and then answer the ques

STALIN [3.7K]

Set up a proportion. 3.50 over 5 = x over 1 (x is what we are trying to find= the cost of 1 lb of apples)
Use cross products and multiply 3.50 by 1 to get 3.50 and then divide that by 5 to get 0.7.

So the apple cost 0.7 or seventy cents per pound.

Check: 0.7*5=3.5

That is correct!

Hope this helps!
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3 0

3 years ago

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Graph the function on the coordinate plane.Type equation here.

gizmo_the_mogwai [7]

Wheres the graph top be able to help?

7 0

2 years ago

Use matrix addition to solve this equation: B + 15 −7 4 0 1 2 = 1 2 12 4 0 2 b11 = b12 = b13 = b21 = 4 b22 = −1 b23 = 0

Arada [10]

Answer:

$b_{11}=-14,b_{12}=9,b_{13}=8,b_{21}=4,b_{22}=-1,b_{23}=0$

Step-by-step explanation:

The given matrix addition is

$B+\begin{bmatrix}15&-7&4\\ 0&1&2\end{bmatrix}=\begin{bmatrix}1&2&12\\ 4&0&2\end{bmatrix}$

We need to find the elements of matrix B.

Let $B=\begin{bmatrix}b_{11}&b_{12}&b_{13}\\ b_{21}&b_{22}&b_{23}\end{bmatrix}$

Substitute the value of matrix.

$\begin{bmatrix}b_{11}&b_{12}&b_{13}\\ b_{21}&b_{22}&b_{23}\end{bmatrix}+\begin{bmatrix}15&-7&4\\ 0&1&2\end{bmatrix}=\begin{bmatrix}1&2&12\\ 4&0&2\end{bmatrix}$