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Burka [1]
3 years ago
15

The absolute value function, f(x) = |x + 2|, is shown.

Mathematics
1 answer:
Anit [1.1K]3 years ago
4 0

Answer:

Incomplete question

Step-by-step explanation:

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Now just 2. Thank you
Levart [38]
The correct choice is B 
x is greater than equals to -4.25
8 0
3 years ago
Graph the linear equation and identify the x intercept <br> Y= 1/5x + 3
KIM [24]

Answer:

x intercept is (-15,0)

Step-by-step explanation:


6 0
3 years ago
Choose the algebraic description that maps the image ΔABC onto ΔA′B′C′.
murzikaleks [220]

The algebraic description that maps the image ΔABC onto ΔA′B′C′ is (x, y) ⇒ (x + 7, y - 4)

<h3>What is transformation?</h3>

Transformation is the movement of a point from its initial location to a new location. Types of transformations are<em> reflection, rotation, translation and dilation.</em>

Translation is the movement of a point either <em>up, left, right or down</em> in the coordinate plane.

The algebraic description that maps the image ΔABC onto ΔA′B′C′ is (x, y) ⇒ (x + 7, y - 4)

Find out more on transformation at: brainly.com/question/4289712

#SPJ1

7 0
2 years ago
I need help with this please and thank you
Kisachek [45]

Answer:

number 2 is SSS

Step-by-step explanation:

6 0
3 years ago
How many feet long is the shadow of a 15-foot flagpole if a 5-foot woman standing at the base of the flagpole has a shadow that
Vikentia [17]

The height of flagpole is 12.5 feet.

Step-by-step explanation:

Given,

Height of woman = 5 foot

Shadow of woman = 6 feet

Ratio of height to shadow = 5:6

Shadow of flagpole = 15 foot

Height of flagpole = x

Ratio of height to shadow = x:15

Using proportion;

Ratio of height to shadow :: Ratio of height to shadow

5:6::x:15\\

Product of mean = Product of extreme

6*x=15*5\\6x=75

Dividing both sides by 6

\frac{6x}{6}=\frac{75}{6}\\x=12.5

The height of flagpole is 12.5 feet.

Keywords: ratio, proportion

Learn more about proportion at:

  • brainly.com/question/9527422
  • brainly.com/question/9381523

#LearnwithBrainly

8 0
3 years ago
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