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GaryK [48]
3 years ago
5

Is the function y= - 9x–6 linear

Mathematics
1 answer:
alina1380 [7]3 years ago
5 0

Answer:

yes

Step-by-step explanation:

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Students love to put ranch dressing on everything, so Elvira needs to keep plenty in stock. The students eat about 2.25 gallons
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In 48 days, they would have consumed 108 gallons, and have 22 left, so yeah. 48 days
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4 years ago
Which graph shows the solution set of the compound inequality or 1.5x-1 > 6.5 or 7x+3 < -25 ?
Dmitry [639]
1.
<span>1.5x-1 > 6.5 
1.5x>1+6.5
</span>1.5x>7.5, divide by 1.5

x>5, is represented by the region to the right of the vertical line x=5 

2. 

<span>7x+3 < -25
7x<-25-3
7x<-28, divide by 4:

x<-4

</span>x<-4, is represented by the region to the left of the vertical line x=-4

Answer: check the picture

6 0
3 years ago
Read 2 more answers
Simplify using appropriate property: 5/7 X 1/5 + 5/7 X 3/5Simplify using appropriate property:
chubhunter [2.5K]

Answer:

4/7

Step-by-step explanation:

5/7 x (1/5+3/5) = 5/7 x 4/5 = 4/7

8 0
3 years ago
What number goes in the highlighted box?
pentagon [3]
-15+7=-8
Adding numbers with different signs makes it subtraction and the answer will have the sign of the bigger number
3 0
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.
4 0
3 years ago
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