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Alex787 [66]
3 years ago
8

Write an equation in point -slope form for the line through the given point with the given slope

Mathematics
1 answer:
lord [1]3 years ago
7 0

Answer: C

Step-by-step explanation:

Using the point-slope form formula, y - y1 = m(x - x1), you would plug in the slope and the points.

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Write the equation for the red part of the piecewise function. Ex: f(x) = mx+b
Akimi4 [234]
The easiest way is to compare two different whole-number points. So, the first one is (0,2) and the other is (3,3), meaning that it's 3 right, 1 up. This means that the slope is ⅓. (⅓x)
Next is the point that intercepts the y-axis. In this case, at x=0 y=2. (+2)

So, the equation for the red part is ⅓x+2
3 0
3 years ago
The scale on a map shows as 1:5000000. What would be the length on the map if the length of a road in
SVEN [57.7K]

Answer:

70 cm

Step-by-step explanation:

\frac{1}{5 * 10^{6} }  = \frac{d}{350} \\\\d = \frac{350}{5 * 10^{6} } \\\\d = 7 * 10^{-5} km = 7 * 10^{-5} * 10^{6} cm = 70 cm

6 0
3 years ago
this is a clear picture from the last one. can u show me the steps of how u worked out the answer plz
azamat

One piece at a time:

4(p+2) means 4 times everything inside the parentheses.
That's 
               4(p)  +  4(2)  =  4p + 8 .

3(2p-3) means 3 times everything inside the parentheses.
That's   
               3(2p)  +  3(-3)  =  6p - 9 .

Now add everything up:

               4p + 8 + 6p - 9 .

Add all the p's together:   4p + 6p = 10p

Add all the plain numbers together:   8 - 9 = -1

The final result is:        10p - 1 .

4 0
3 years ago
Becca has $18.69 for a buffet. There is an entrance fee of $3.09 with a $3.62 charge for every pound of crab legs ordered. This
s2008m [1.1K]
3.62p+3.09=18.69
-3.09 -3.09
——————————-
3.62p = 15.60
15.60/3.62
P= 4.31
She can by 4.31 pounds
7 0
4 years ago
What is the range of the function y = x 2?
Leno4ka [110]

Answer:

\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}

The graph is also attached below.

Step-by-step explanation:

Given the function

y=x^2

  • We know that the range of a function is the set of values of the dependent variable for which a function is defined.

\mathrm{For\:a\:parabola}\:ax^2+bx+c\:\mathrm{with\:Vertex}\:\left(x_v,\:y_v\right)

\mathrm{If}\:a

\mathrm{If}\:a>0\:\mathrm{the\:range\:is}\:f\left(x\right)\ge \:y_v

a=1,\:\mathrm{Vertex}\:\left(x_v,\:y_v\right)=\left(0,\:0\right)

f\left(x\right)\ge \:0

Thus,

\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}

The graph is also attached below.

5 0
3 years ago
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