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3 years ago

7

Write a function rule for "The output is 5 less than four times the input." Use x for the independent variable and y for the dep

endent variable.
y = ?

1 answer:

Elza [17]3 years ago

6 0

Answer:

32

Step-by-step explanation:

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Answer it answer it answer it

AysviL [449]

Answer: D.

Step-by-step explanation: If you look closely, putting a 100 wouldn't make sense

6 0

3 years ago

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I need help with answering the following questions. :/ :)

s344n2d4d5 [400]

Answer:

a) (0, ∞)

b) (-∞, ∞)

c) x = 0

Step-by-step explanation:

It helps to have some idea what the log function is.

__

a) The domain is all positive numbers: (0, ∞).

b) The range is all real numbers: (-∞, ∞). (The vertical translation downward by 5 units does not change that.)

c) There is a vertical asymptote where the argument of the log function is zero: at x=0.

3 0

3 years ago

Find the probability on #4<br> Please

julia-pushkina [17]

The probability of drawing a 5 from card numbered 1 - 10 is 1/10 because only one card out of ten is numbered 5.
The probability of rolling a 2 on the cube is 1/6 because there are 6 faces of the cube with 6 different numbers. The probability of picking a 5 and rolling a 2 together is 1/60 because
1/10 x 1/6 = 1/60. You multiply them together because there will be 60 combinations: 10 card possibilities and 6 cube possibilities.
Hope this helps!

3 0

4 years ago

The midpoint of AB is at (3,7) and A is at (0,-5). Where is B located?

olga nikolaevna [1]

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ \square }}\quad ,&{{ \square }})\quad 
% (c,d)
&({{ \square }}\quad ,&{{ \square }})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)\qquad thus
\\
----------------------------\\$ $\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ 0}}\quad ,&{{ -5}})\quad 
% (c,d)
B&({{ \square }}\quad ,&{{ \square }})
\end{array}\qquad
% coordinates of midpoint 
(3,7)\impliedby midpoint\qquad thus
\\ \quad \\
\left(\cfrac{{{ x_2 }} + {{ 0}}}{2}=3\quad ,\quad \cfrac{{{ y_2 }} + {{( -5)}}}{2}=7 \right)\to 
\begin{cases}
\cfrac{{{ x_2 }} + {{ 0}}}{2}=3
\\ \quad \\
\cfrac{{{ y_2 }} + {{ -5}}}{2}=7
\end{cases}
\\ \quad \\
solve\ for\ x_2\ and\ y_2$

3 0

3 years ago

ANSWER THIS CORRECTLY AND I WILL 5 STAR U AND GIVE U POINTS AND THANK YOU

Ivanshal [37]

Answer:

C

Step-by-step explanation:

given:

m= -2 and (3, -1)

point-slope form: y-y1= m(x-x1)

y - (-1) = -2(x-3)

y + 1 = -2(x-3)

7 0

3 years ago

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