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2 years ago

Help me with this pleaseeeee

Mathematics

2 answers:

tangare [24]2 years ago

6 0

Answer:

i notice that she only has $30 to spend for a month. and she has to pay $21 every month including $3 for every download so she can only download 3 1GB of data

Step-by-step explanation:

creativ13 [48]2 years ago

3 0

3 GBs per month because you have to take 30-21 (the monthly cost of the phone)=9 and each GB is $3 a month so 9 divided by 3= 3

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Write the equation of a line that is perpendicular to the line that passes through (4,−3)and (−7,8).

masha68 [24]

Line perpendicular to a given line is = 1/m

To find m, use m =y2-y1/x2-x1 formula

m = (8-(-3))/(-7-4)

m = 11/-11,

m = -1

Use y= mx+c to find equation of line

Plug in a pair of values,

-3= -1(4)+ c

c= 1

Thus, equation is:

y = -x+1

Equation of line perpendicular to this line is:

y= x+1

Hope it helps :)

Pls mark it as Brainliest :)

3 0

3 years ago

PLZ help X=????????????????????????

Leya [2.2K]

Answer:

x=1

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

8 0

3 years ago

Solve the equation 6-x-x=18 A. -6 B. 6 C. 9 D. -9

Veronika [31]

A. x = -6
6-x-x = 18
6-2x =18
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x = -6

4 0

3 years ago

A cyclist travels at distance of 400 meters in 120 seconds towards school, calculate his speed. (Show your work)

DochEvi [55]

we know that

The scalar magnitude of the velocity vector is the speed. The speed is equal to

$Speed=\frac{distance}{time}$

in this problem we have

$distance=400\ m \\time=120\ sec$

substitute in the formula

$Speed=\frac{400}{120}$

$Speed=3.5\frac{m}{sec}$

therefore

the answer Part a) is

the speed is equal to $3.5\frac{m}{sec}$

Part b) Find the velocity

we know that

Velocity is a vector quantity; both magnitude and direction are needed to define it

in this problem we have

the magnitude is equal to the speed

$magnitude=3.5\frac{m}{sec}$

$direction=North\ East\ (NE)$

therefore

the answer Part b) is

the velocity is $3.5\frac{m}{sec}\ North\ East\ (NE)$

Part c)

we know that

the acceleration is equal to the formula

$a=\frac{V2-V1}{t2-t1}$

in this problem we have

$V2=0 \\V1=3.5\frac{m}{sec}$

$t2=15\ sec\\t1=0$

substitute in the formula

$a=\frac{0-3.5}{15-0}$

$a=-\frac{3.5}{15}\frac{m}{sec^{2}}$

$a=-\frac{7}{30}\frac{m}{sec^{2}}$

therefore

the answer Part c) is

the acceleration is $-\frac{7}{30}\frac{m}{sec^{2}}$

This is an example of negative acceleration

7 0

3 years ago

Read 2 more answers

Is this answer correct??

Sauron [17]

Yes its correct. 7.4-.2=7.2

3 0

3 years ago