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2 years ago

PLEASE HELP ASAP!!!!!! WILL GIVE BRAINLEST!!!!!

Mathematics

1 answer:

OLga [1]2 years ago

6 0

I think the answer is A
hope i helped

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Please help me with precal

Doss [256]

Consider the function $f(x)=\sqrt{x}.$ This function has:

the domain $x\in [0,\infty);$
the range $y\in [0,\infty).$

If the domain of unknown function is $[a,\infty),$ then $x\ge a$ or $x-a\ge 0.$ This means that you have $\sqrt{x-a}$ as a part of needed function.

If the range of unknown function is $[b,\infty),$ then $y\ge b.$ This means that you have to translate function b units up and then the expression of the function is

$y=\sqrt{x-a}+b.$

Answer: correct choice is B.

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2 years ago

A shopper keeper sold a total of 15 boxes of pencils on Monday and Tuesday. She still three more boxes on Monday then on Tuesday

Nookie1986 [14]

I think 180 that should be your answer sorry i did not get here quick enough

8 0

2 years ago

4 + 2x2(3x-5)<br> number of terms

igor_vitrenko [27]

3, assuming 2x2 is 2x squared, u will get
4 + 6x3 - 10x2

So 3 terms

7 0

2 years ago

Simplify 4 over 7 divided by 3 over negative 8

Stolb23 [73]

-1 5/11

Step-by-step explanation:

Dividing fractions is easy as pie, flip the second one over then multiply

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3 years ago

Polygon PQRS, shown in the figure, is dilated by a scale factor of 0.5 with the origin as the center of dilation, resulting in p

nata0808 [166]

Based on the above:

The slope of P'Q' is = -3/2

The length of P'Q' is approximately = $\sqrt[3]{13}$

<h3>What is the polygon about?</h3>

The slope of P'Q'

= -6/4 = -3/2

The length of P'Q' =

P'Q' = $\sqrt{6^2 + 4^2}$

= $\sqrt{52}$

= $\sqrt[2]{13}$

Therefore;

P'Q' = $\frac{3}{2}$

P'Q' = $\frac{3}{2}$ x $\sqrt[2]{13}$

= $\sqrt[3]{13}$

Learn more about polygon from

brainly.com/question/1592456

#SPJ1

7 0

1 year ago