A mechanic charges $75 per hour for her services and $60 for the initial call. What is the dependent variable in the situation?

Mathematics

1 answer:

tresset_1 [31]3 years ago

8 0

Answer:

The amount of hours it takes for the service to be completed

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What is the area of the triangle shown below?<br> 14

Sveta_85 [38]

Answer:

Excuse me but they is nothing shown below!

Step-by-step explanation: Again not trying to be rude but there is nothing shown below! thank you for your time! <3

5 0

3 years ago

Please help it’s the last question

Mrac [35]

Answer:

$y=$ $-\frac{5}{6}x - 1$ is the answer

Step-by-step explanation:

Equation of the line: y = 6/5x + 1

= 5y = 6x + 5

= 6x - 5y + 5

Equation of the perpendicular line: bx - ay + k = 0

= -5x -6y + k = 0

Equation passes through (6,-6),

-5(6) -6(-6) + k = 0

-30 + 36 + k = 0

6 + k = 0

k = -6

Substituting,

-5x -6y + k = 0

-5x -6y -6 = 0

-6y = 5x + 6

$y=$ $-\frac{5}{6}x - 1$ (Slope-Intercept form)

6 0

3 years ago

The graph of F(x) shown below has the same shape as the graph of G(x) = x^2 but it is shifted down 5 units and to the left 4 uni

grandymaker [24]

$\bf ~\hspace{10em}\textit{function transformations} \\\\\\ \begin{array}{llll} f(x)= A( Bx+ C)^2+ D \\\\ f(x)= A\sqrt{ Bx+ C}+ D \\\\ f(x)= A(\mathbb{R})^{ Bx+ C}+ D \end{array}\qquad \qquad \begin{array}{llll} f(x)=\cfrac{1}{A(Bx+C)}+D \\\\\\ f(x)= A sin\left( B x+ C \right)+ D \end{array} \\\\[-0.35em] ~\dotfill\\\\ \bullet \textit{ stretches or shrinks horizontally by } A\cdot B\\\\ \bullet \textit{ flips it upside-down if } A\textit{ is negative}\\ ~~~~~~\textit{reflection over the x-axis}$

$\bf \bullet \textit{ flips it sideways if } B\textit{ is negative}\\ ~~~~~~\textit{reflection over the y-axis} \\\\ \bullet \textit{ horizontal shift by }\frac{ C}{ B}\\ ~~~~~~if\ \frac{ C}{ B}\textit{ is negative, to the right}\\\\ ~~~~~~if\ \frac{ C}{ B}\textit{ is positive, to the left}\\\\ \bullet \textit{ vertical shift by } D\\ ~~~~~~if\ D\textit{ is negative, downwards}\\\\ ~~~~~~if\ D\textit{ is positive, upwards}\\\\ \bullet \textit{ period of }\frac{2\pi }{ B}$

with that template in mind, let's see

down by 5 units, D = -5

to the left by 4 units, C = +4

$\bf G(x)=x^2\implies G(x)=1(1x+\stackrel{C}{0})^2+\stackrel{D}{0} \\\\\\ \begin{cases} D=-5\\ C=+4 \end{cases}\implies F(x)=1(1x+\stackrel{C}{4})^2\stackrel{D}{-5}\implies F(x)=(x+4)^2-5$