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Airida [17]
3 years ago
14

Which Of the following is the value of 7-15:3-2?​

Mathematics
1 answer:
Tanzania [10]3 years ago
5 0

Answer:

-8:1

Step-by-step explanation:

How to solve the function operation :D

Simplify 7-15= -8

Then simplify 3-2= 1

Total= -8:1

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Ty has 230 pages of text to read in the next three days. if he averages 25 pages every 1 1/4 hours, how many hours is he going t
sineoko [7]

Answer:

11.5 hours

Step-by-step explanation:

25 pages every 75 minutes (1 1/4)

Thats means 1 page every 3 minutes (divide 25 and 75 by 25 to get this)

230 pages would mean 690 minutes (times the 3 by 230)

690/6 = 11.5

7 0
3 years ago
Using the Slope-Intercept form, find the equation of a line that has a slope of 3 and y - intercept (2)
RideAnS [48]

Answer:

y = 3x + 2

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here m = 3 and c = 2 , then

y = 3x + 2 ← equation of line

5 0
3 years ago
If you vertically stretch the exponential function f(x) = 2^x by a factor of 5, what is the equation of the new function?
lara31 [8.8K]

Answer:

B

Step-by-step explanation:

In this case we multiply the function f(x) = 2^x by 5, so the correct answer is B.

6 0
4 years ago
Write a two column proof 2x+6 and 96 prove: x=45
viktelen [127]
2x+6=96
2(45)+6
90+6=96
96=96
hope this helps u out!
6 0
3 years ago
ITS URGENT PLEASE HELP Which equation represents the line that is parallel to y=3/7x+11 and passes through (-21,42)?
Pepsi [2]

Answer:

y = 3/7x + 51

Step-by-step explanation:

y = 3/7x + 11 is parallel to the line so it means that both of their gradient are same so:

y = mx + c

m= gradient point = (-21 , 42)

= 3/7

now we sub the m and the point into the formula which is y= mx+c because we should find the c first then we can find the equation of the line:

42 = 3/7 (-21) + c

42 = -9 + c

42 + 9 = c

51 = c

c = 51

now you hv to rewrite again the equation become y= 3/7x + 51

So now your final answer is : y = 3/7x +51

3 0
3 years ago
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