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coldgirl [10]
3 years ago
13

How do I solve a+18=7a ?

Mathematics
2 answers:
Andre45 [30]3 years ago
7 0
a+18=7a

18=7a-a

18=6a ⇒ a= \frac{18}{6}

\boxed {a=3}
Slav-nsk [51]3 years ago
6 0
a+18=7a \ \ \ |\hbox{subtract 7a from both sides} \\
-6a+18=0 \ \ \ |\hbox{subtract 18 from both sides} \\
-6a=-18 \ \ \ |\hbox{divide both sides by (-6)} \\
a=3
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A colored ball was drawn out of a bag with replacement. What is
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Answer:

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A rectangular room is 12.5 feet wide and 14 feet long. How many linear feet of baseboards are needed to trim the room?
QveST [7]

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8 0
4 years ago
(-6,8); perpendicular to y = -3/2x -1
UNO [17]

Answer:

y =  \frac{2}{3} x + 12

Step-by-step explanation:

y =  -  \frac{3}{2} x - 1

The gradient of a line is the coefficient of x when the equation of the line is written in the form of y=mx+c.

Thus, gradient of given line=-  \frac{3}{2}.

The product of the gradients of perpendicular lines is -1.

(Gradient of line)(-3/2)= -1

Gradient of line

- 1 \div ( -  \frac{3}{2} ) \\  =  - 1( -  \frac{2}{3} )  \\  =  \frac{2}{3}

Substitute m=\frac{2}{3} into y=mx+c:

y =  \frac{2}{3} x + c

To find the value of c, substitute a pair of coordinates.

When x= -6, y= 8,

8 =  \frac{2}{3} ( - 6) + c \\  \\ 8 =  - 4 + c \\ c = 8 + 4 \\ c = 12

Thus, the equation of the line is y =  \frac{2}{3} x + 12.

7 0
3 years ago
Find the solution of the given initial value problems in explicit form. Determine the interval where the solutions are defined.
natali 33 [55]

Answer:

The solution of the given initial value problems in explicit form is y=x-x^2-2  and the solutions are defined for all real numbers.

Step-by-step explanation:

The given differential equation is

y'=1-2x

It can be written as

\frac{dy}{dx}=1-2x

Use variable separable method to solve this differential equation.

dy=(1-2x)dx

Integrate both the sides.

\int dy=\int (1-2x)dx

y=x-2(\frac{x^2}{2})+C                  [\because \int x^n=\frac{x^{n+1}}{n+1}]

y=x-x^2+C              ... (1)

It is given that y(1) = -2. Substitute x=1 and y=-2 to find the value of C.

-2=1-(1)^2+C

-2=1-1+C

-2=C

The value of C is -2. Substitute C=-2 in equation (1).

y=x-x^2-2

Therefore the solution of the given initial value problems in explicit form is y=x-x^2-2 .

The solution is quadratic function, so it is defined for all real values.

Therefore the solutions are defined for all real numbers.

4 0
3 years ago
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