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Ymorist [56]
3 years ago
8

What’s the correct answer for this question

Mathematics
1 answer:
HACTEHA [7]3 years ago
6 0

All of them are true except AD = GH.

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If g(x) = 4√x-8 , which statement is true?
Mariana [72]
We have the following function:

g(x)=4\sqrt{x}-8

So if we graph this function we will get the Figure below. Thus, let's study both the equation and the graph to get some conclusions. Therefore, we can assure these statements:

First. The function is defined only for x\geq 0 as shown in the Figure. This is also true because of \sqrt{x} where x must be greater (or equal) than zero.

Second. The range of the function are the values of y\geq -8.

Third. If x creases then y always creases, too. 

4 0
3 years ago
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How many solutions does the equation below have?
Elina [12.6K]

Answer:

1.one solution is the answer

Step-by-step explanation:

3 0
2 years ago
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Don’t stress. try taking a nap. or watching a movie. Or listen to calming music.
4 0
3 years ago
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Pleaseeee helpppp!!!!​
kondor19780726 [428]

The right answer is Option A:x=\frac{rs-t}{2}

Step-by-step explanation:

Given expression is;

s=\frac{2x+t}{r}

We will solve this expression for x, therefore,

Multiplying both sides by r

r*s=\frac{2x+t}{r}*r\\rs=2x+t

Subtracting t from both sides

rs-t=2x+t-t\\rs-t=2x\\2x=rs-t

Dividing both sides by 2

\frac{2x}{2}=\frac{rs-t}{2}\\x=\frac{rs-t}{2}

The right answer is Option A:x=\frac{rs-t}{2}

Keywords: subtraction, division

Learn more about subtraction at:

  • brainly.com/question/1238144
  • brainly.com/question/12497249

#LearnwithBrainly

7 0
3 years ago
Points M, N, and P are respectively the midpoints of sides AC , BC , and AB of △ABC. Prove that the area of △MNP is on fourth of
Hunter-Best [27]

Answer:

The area of △MNP is one fourth of the area of △ABC.

Step-by-step explanation:

It is given that the points M, N, and P are the midpoints of sides AC, BC and AB respectively. It means AC, BC and AB are median of the triangle ABC.

Median divides the area of a triangle in two equal parts.

Since the points M, N, and P are the midpoints of sides AC, BC and AB respectively, therefore MN, NP and MP are midsegments of the triangle.

Midsegments are the line segment which are connecting the midpoints of tro sides and parallel to third side. According to midpoint theorem the length of midsegment is half of length of third side.

Since MN, NP and MP are midsegments of the triangle, therefore the length of these sides are half of AB, AC and BC respectively. In triangle ABC and MNP corresponding side are proportional.

\triangle ABC \sim \triangle NMP

MP\parallel BC

MP=\frac{BC}{2}

By the property of similar triangles,

\frac{\text{Area of }\triangle MNP}{\text{Area of }\triangle ABC}=\frac{PM^2}{BC^2}

\frac{\text{Area of }\triangle MNP}{\text{Area of }\triangle ABC}=\frac{(\frac{BC}{2})^2}{BC^2}

\frac{\text{Area of }\triangle MNP}{\text{Area of }\triangle ABC}=\frac{1}{4}

Hence proved.

5 0
3 years ago
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