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3 years ago

What is the value of x

Mathematics

2 answers:

faust18 [17]3 years ago

7 0

X=6 is the answer sorry if I’m wrong

Lelu [443]3 years ago

5 0

Answer:

x=6

Step-by-step explanation:

Because these lines are parallel, that means these angles are congruent, so you can set up an equation because they are equal to each other.

110=19x-4

add 4 to both sides

114=19x

and divide by 19

6=x

x=6

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Oksana_A [137]

Answer:

a) $y=8x-1000$

b) $profit=\$8$

c) They'd have lost $1000 if they had sold no calendars.

Step-by-step explanation:

a) The equation of the line in Slope-Intercept form is:

$y=mx+b$

Where "m" is the slope and "b" is the y-intercept.

In this case we know that "y" represents the profit of loss and "x" the number of calendars sold.

Then, according to the exercise, the line passes through these two points:

$(80,-360)$ and $(200,600)$

Then, we can find the slope of the line with the formula $m=\frac{y_2-y_1}{x_2-x_1}$

$m=\frac{-360-600}{80-200}=8$

Now, we can substitute the slope and one of those points into $y=mx+b$ and solve for "b":

$600=8(200)+b\\\\b=-1000$

Then, subtituting values, we get that the equation that describes the relation between the profit of loss and the number of calendars sold, is:

$y=8x-1000$

b) The slope of the line is the profit they made from selling each calendar

$profit=8$

c) The y-intercept is the amount they would have lost if they had sold no calendars:

$b=-1000$

They'd have lost $1000 if they had sold no calendars.

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3 years ago

What is the value of m, when 2⁶ x 4 = 418

yaroslaw [1]

Answer:

The answer should be 3.3/4

Step-by-step explanation:

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2 years ago

Suppose the man in the St. Ives poem has x wives, each wife has x sacks, each sack has x cats, and each cat has x kits. Write an

Harrizon [31]

Answer:

$x^{1}$ wives

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Step-by-step explanation:

$x^{1}$ wives

If each of the "x" wives has "x" sacks, so the number of sacks is:

$x^{2}$ sacks

If each of the "x" wives has "x" sacks, and each sack has "x" cats, so the number of cats is:

$x^{3}$ cats

If each of the "x" wives has "x" sacks, and each sack has "x" cats, and each cat has "x" kits, so the number of kits is:

$x^{4}$ kits

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