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exis [7]
3 years ago
5

Twice the sum of a number and 7 is equal to three times the difference of the number and 8. Find the number.

Mathematics
1 answer:
amm18123 years ago
3 0

Answer:

the number is 10.

Step-by-step explanation:

2(x + 7) = 3(x - 8)

2x + 14 = 3x - 24

2x = 3x -10

x = 10

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If f(x) = 4x - 7, what's the value of f(2) ? *<br> O -1<br> O 1<br> O 20<br> O 35
sp2606 [1]

Answer:

B

Step-by-step explanation:

f(x) = 4x - 7

What does f(x) mean? It means that whatever variable you have on the left which is (x), the same variable must be on the right.

So f(2) means that whatever variable is on the right, it must be replaced with 2 on the right. So .....

f(2) = 4(2) - 7

f(2) = 8 - 7

f(2) = 1

3 0
2 years ago
6=−w8 how can i solve this
Vaselesa [24]
Divide 6 by -8 to get the variable alone!
6/-8 < division< equation
4 0
3 years ago
An ice cream shop sold 12,789 chocolate and 9,324 cookie dough cones. They sold 1,078 more peanut butter cones than cookie dough
juin [17]
12,789 chocolate
9,324 Cookie Dough
10,402 Peanut Butter Cones
13,788 Vanilla

46,303 in total

3 0
2 years ago
Let f(x) = 7x − 13. Find f−1(x)
motikmotik
f(x)=7x-13\\\\y=7x-13\ \ \ \ |add\ 13\ to\ both\ sides\\\\y+13=7x\ \ \ \ \ |divide\ both\ sides\ by\ 7\\\\x=\dfrac{y+13}{7}\\\\Answer:f^{-1}(x)=\dfrac{y+13}{7}
6 0
3 years ago
Here are the endpoints of the segments BC, FG, and JK.<br> B, −67
yulyashka [42]

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{-6}~,~\stackrel{y_1}{7})\qquad C(\stackrel{x_2}{-4}~,~\stackrel{y_2}{4})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ BC=\sqrt{[-4 - (-6)]^2 + [4 - 7]^2}\implies BC=\sqrt{(-4+6)^2+(-3)^2} \\\\\\ BC=\sqrt{2^2+(-3)^2}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{distance between 2 points}

F(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-4})\qquad G(\stackrel{x_2}{1}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ FG=\sqrt{[1 - (-2)]^2 + [-2 - (-4)]^2}\implies FG=\sqrt{(1+2)^2+(-2+4)^2} \\\\\\ FG=\sqrt{9+4}\implies \boxed{FG=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ J(\stackrel{x_1}{4}~,~\stackrel{y_1}{2})\qquad K(\stackrel{x_2}{5}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}

JK=\sqrt{[5 - 4]^2 + [-2 - 2]^2}\implies JK=\sqrt{1^2+(-4)^2}\implies \boxed{JK=\sqrt{17}} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \overline{BC}\cong \overline{FG}~\hfill

4 0
1 year ago
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