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3 years ago

What’s the answer ????

Mathematics

2 answers:

Citrus2011 [14]3 years ago

7 0

Answer:

C. The graph would shift 5 units to the left

Step-by-step explanation:

If (x+5) was substituted in the place of x then it would shift 5 units to the left

Answer is C.

Katen [24]3 years ago

6 0

Answer:

C. The graph would shift five units to the left.

Step-by-step explanation:

y = x² + 5x - 2

We can write this equation in the form

y = a(x - h)² + k

If you change the value of h, you are <em>translating the parabola horizontally</em>.

If you <em>increase h by five units, the graph shifts five units to the left</em>.

In the figure below, the red parabola represents the original equation.

The blue parabola shows the graph shifting five units to the left

(from x = -2.5 to x = -7.5) when you replace x by x + 5.

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pishuonlain [190]

The commitment adherence percentage for the week will be 53%.

<h3>How to compute the percentage?</h3>

It was stated that Amy was scheduled to service 17 intervals this week. She picked up 4 more intervals during the week.

The following can be deduced based on the information that are given:

Total intervals = 17

Taken intervals = 4+ 3 + 2 = 9

Therefore, the commitment adherence percentage for the week will be:

= 9/17 × 100

= 53%

Learn more about percentages on:

brainly.com/question/24304697

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1 year ago

Each side of a square field is 13 m. Find the distance covered by a man going around the field five times

strojnjashka [21]

Step-by-step explanation:

going around the field one time is walking all 4 sides.

so, 4×13 m

going around the field 5 times is walking all 4 sides 5 times.

that means

5×4×13 m = 20×13 = 260 m

8 0

3 years ago

What the perimeter of the triangle

Akimi4 [234]

Answer:

Step-by-step explanation:

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4 0

3 years ago

Read 2 more answers

Can you please find the missing side these are similar figures.

storchak [24]

Answer:

1.108

2.4

3.6

4.28

5.80

6. 10.5

hope this will help you

8 0

3 years ago

Evaluate 2/3 + 1/3 + 1/6 + … THIS IS CONTINUOUS. It is NOT as simple as 2/3 + 1/3 + 1/6.

Vlada [557]

$a=\dfrac{2}{3}\\r=\dfrac{1}{2}$

The sum exists if $|r|$

$\left|\dfrac{1}{2}\right|$ therefore the sum exists

$\displaystyle\\\sum_{k=0}^{\infty}ar^k=\dfrac{a}{1-r}$

$\dfrac{2}{3}+\dfrac{1}{3}+\dfrac{1}{6}+\ldots=\dfrac{\dfrac{2}{3}}{1-\dfrac{1}{2}}=\dfrac{\dfrac{2}{3}}{\dfrac{1}{2}}=\dfrac{2}{3}\cdot 2=\dfrac{4}{3}$

5 0

3 years ago