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3 years ago

What is an equation of the line that passes through the points (-8,4) and (-8, 8)?

Mathematics

1 answer:

grin007 [14]3 years ago

3 0

Answer:

X= -8

Step-by-step explanation:

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Name three decimals with a product that is about 40

yulyashka [42]

4.0
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.040
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3 years ago

I need the work to please

Free_Kalibri [48]

There you go I hope it helps

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3 years ago

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there are 20 pieces of candy in a bag. 12 pieces are chocolate. what percentage of the pieces in the bag are chocolate

gavmur [86]

Answer:

Step-by-step explanation:

12/20 = 0.6

0.6x100=60%

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3 years ago

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I need help with making predictions with probability please help me correctly answer these questions.

Semmy [17]

Answer:

1. So the porpotion should be 25/100 = x/120

You have this because you know that 25% of the dogs who go to ABC veterinarian need a rabies booster. So 100% of the dog should be 120 dogs, and now our job is to find how many dogs make up 25% percent of the total number.

According to our porpotion, we have:

(120 . 25)/100 = 30 (dogs)

Another way of understanding this is that you have to find 25% of the total amount, so just take that total number and multiply it with 25% or 25/100 which is 0.25 as well, that's why we multiply it with 25 first and then divide it to 100 later.

*I guess there is no need of explaining it

0.9 × 550 = 495

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3 years ago

Determine whether the set of vectors <img src="https://tex.z-dn.net/?f=%20v_%7B1%3D%283%2C2%2C1%29%2C%20v_%7B2%7D%20%3D%28-1%2C-

Korolek [52]

Since each vector is a member of $\mathbb R^3$ , the vectors will span $\mathbb R^3$ if they form a basis for $\mathbb R^3$ , which requires that they be linearly independent of one another.

To show this, you have to establish that the only linear combination of the three vectors $c_1\mathbf v_1+c_2\mathbf v_2+c_3\mathbf v_3$ that gives the zero vector $\mathbf0$ occurs for scalars $c_1=c_2=c_3=0$ .

$c_1\begin{bmatrix}3\\2\\1\end{bmatrix}+c_2\begin{bmatrix}-1\\-2\\-4\end{bmatrix}+c_3\begin{bmatrix}1\\1\\-1\end{bmatrix}=(0,0,0)\iff\begin{bmatrix}3&-1&1\\2&-2&1\\1&-4&-1\end{bmatrix}\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$

Solving this, you'll find that $c_1=c_2=c_3=0$ , so the vectors are indeed linearly independent, thus forming a basis for $\mathbb R^3$ and therefore they must span $\mathbb R^3$ .

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3 years ago