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zmey [24]
3 years ago
7

Goven the graphs of h(x) and k(x) find:

Mathematics
2 answers:
BabaBlast [244]3 years ago
4 0

i think the answer is b

sergejj [24]3 years ago
4 0

Answer:

I believe the answer would be B.

Step-by-step explanation:

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Solve the equation 3(2x + 2) = 3x − 15.
Margaret [11]

Hi there! :)

Answer:

x = -7.

Step-by-step explanation:

Starting with:

3(2x + 2) = 3x - 15

Begin by distributing '3' with the terms inside of the parenthesis:

3(2x) + 3(2) = 3x - 15

Simplify:

6x + 6 = 3x - 15

Isolate the variable by subtracting '3x' from both sides:

6x - 3x + 6 = 3x - 3x - 15

3x + 6 = -15

Subtract 6 from both sides:

3x + 6 - 6 = -15 - 6

3x = -21

Divide both sides by 3:

3x/3 = -21/3

x = -7.

6 0
3 years ago
Read 2 more answers
Question: In Boston, Massachusetts, there is a 60% probability of rain on a specific day. How was this probability most likely d
alex41 [277]

Answer:

On days with similar conditions as this day, it has rained 60% of the time in Boston, Massachusetts

Step-by-step explanation:

I got it right on the quiz

6 0
2 years ago
A 13 Ft ladder is leaning up against a building, with the base of the ladder 2 feet from the building. How high up on the buildi
Lelechka [254]

Visualize the image in your mind and you can see that it forms a right angled triangle with a base,altitude and hypotenuse

base=2 feet

hypotenuse(ladder)=13 ft

We need to find the altitude,

By applying pythagoras theorem,

base square+altitude square=hypotenuse square

Take altitude as x

2 square+x square=13 square

4+x square=169

x square=169-4

x square=165

x= root of 165

x=12.8 feet(approx.)

6 0
3 years ago
Find the mean median mode and range 2,5,9,4,3
snow_lady [41]
Mean: 4.6 (basically the average
Median: 4 (the number in the middle
Mode: there isn’t one (it’s the number repeated the most)
Range: 7 (largest and smallest number subtracted
3 0
3 years ago
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PLEASE HELP ME!!! <br> Solve in substitution form step by step please
romanna [79]

Answer:

3. (3,-4)

Step-by-step explanation:

8 0
2 years ago
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