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2 years ago

What is the equation of a horizontal line passing through the point (−7, 5)? (1 point)

1 answer:

6 0

If it’s a horizontal line, we focus on the y-coordinates.

(-7,5)

-7 is a x-coordinate
5 is a y-coordinate

The answer is y = 5

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HELP PLEASEEEEEEEEEE

defon

Answer:

x=8

Step-by-step explanation:

because x is a pronoun in the laws of math. Hope this helps! lol

5 0

2 years ago

HELP ASAP

vovangra [49]

Answer:

See below and the attachment

Step-by-step explanation:

Refer to attached graph.

It includes the points and a best fit line produced by a graphing calculator.

17. The scatter plot is attached
18. There is a positive slope. Number of field goal attempts increase as minutes played increase.
19. The point is plotted: 2500 minutes gives about 880 field goal attempts.

4 0

2 years ago

Answer Is not C please help

Katyanochek1 [597]

Answer:

Step-by-step explanation:

Well since it is arthimetic sequence we know the formula

We know d is difference between second term and first so it will be -5- (-12)= -5+12= 7

$a_{n}= a _{1}+ d (n-1) \\a_{n}= -12+7(n-1)\\a_{n} = -12+7n-7\\a_{n}= 7n-19$

the domain of this is 1,2,3,4.......

range is -12,-5,2,9.......

Therefore answer is B.

5 0

2 years ago

Given the vector (4|3) and the transformation matrix (0|1|-1|0), which vector is the imagine after applying the transformation t

horrorfan [7]

Answer:

C.(3|-4)

Step-by-step explanation:

Given the vector:

$\left[\begin{array}{ccc}4\\3\end{array}\right]$

The transformation Matrix is:

$\left[\begin{array}{ccc}0&1\\-1&0\end{array}\right]$

The image of the vector after applying the transformation will be:

$\left[\begin{array}{ccc}0&1\\-1&0\end{array}\right]\left[\begin{array}{ccc}4\\3\end{array}\right]\\\\=\left[\begin{array}{ccc}0*4+1*3\\-1*4+0*3\end{array}\right]\\\\=\left[\begin{array}{ccc}3\\-4\end{array}\right]$

The correct option is C

7 0

3 years ago

Please round to the nearest tenth if possible.

Butoxors [25]

$\bf \cfrac{64~\underline{mi}}{\underline{hr}}\cdot \cfrac{\underline{hr}}{60~sec}\cdot \cfrac{5200~ft}{\underline{mi}}\implies \cfrac{64\cdot 5200~ft}{60~sec}\implies \cfrac{16640~ft}{3~sec}
\\\\\\
\approx 5546.67\frac{ft}{sec}\\\\
-------------------------------\\\\
\cfrac{6000~\underline{lbs}}{\underline{day}}\cdot \cfrac{ton}{2000~\underline{lbs}}\cdot \cfrac{7~\underline{day}}{week}\implies \cfrac{6000\cdot 7~ton}{2000~week}\implies 21.00\frac{ton}{week}$

$\bf -------------------------------\\\\
\cfrac{2~\underline{lbs}}{\underline{week}}\cdot \cfrac{16~oz}{\underline{lbs}}\cdot \cfrac{\underline{week}}{7~day}\implies \cfrac{2\cdot 16~oz}{7~day}\implies \cfrac{32~oz}{7~day}\approx 4.57\frac{oz}{day}$

8 0

3 years ago