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Scilla [17]
3 years ago
9

About 0.5 of all the students in the fifth grade are girls. If the fifth grade has 80 ​students, how many of the students are gi

rls​?
Mathematics
1 answer:
Nookie1986 [14]3 years ago
7 0

Answer:

40 Girls.

Step-by-step explanation:

80 multipled by 0.5 = 40

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the midpoint of a horizontal line segment is at x=1 if the length of the line segment is 10.5. what absolute value equation coul
egoroff_w [7]

Answer: -4.25 ≤ x ≤ 6.25

Step-by-step explanation:

The midpoint of a segment is located at the same distance of each of the endpoints of the segment.

The midpoint is at x = 1.

The length of the segment is 10.5, if we divide it by two we have:

10.5/2 = 5.25

Now, if we want the endpoints to be at the same distance of the midpoint, then the endpoints will be:

Xmax = midpoint + 5.25

Xmin = midpoint - 5.25

Then the extremes are:

Xmax = 1 + 5.25 = 6.25

Xmin = 1 - 5.25 = -4.25

Then this segment can be written as:

Xmin ≤ x ≤ Xmax

-4.25 ≤ x ≤ 6.25

8 0
3 years ago
Does anyone know this answer
Rzqust [24]

Answer:

c

Step-by-step explanation:

took test

4 0
2 years ago
How much money would need to be deposited into an account earning 5.75% interest compounded annually in order for the accumulate
Leona [35]
I will assume you are using compound interest. 

<span>let the amount invested be x </span>

<span>x(1.0575)^25 = 85000 </span>
<span>x = 85000/1.0575^25 = $21,009.20</span>
6 0
3 years ago
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Do any of you have a made-up pokemon? Give Me Name, Hight, Type, Category, And All of That Stuff.
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Answer:

I like pikachu. I don't know how to do te rest though

Step-by-step explanation:

7 0
3 years ago
Please help me with this
Ne4ueva [31]

Answer:

Option c

Step-by-step explanation:

A system of equations are given to us. And we need to solve them . The given system is

\begin{cases} y = 2x - 3\dots 1 \\ y = x^2 - 3\dots 2 \end{cases}

We numbered the equations here . Now put the value of equation 1 in equation 2 that is substituting y = 2x - 3 in eq. 2 .

\implies y = x^2 - 3 \\\\\implies 2x - 3 = x^2 - 3 \\\\\implies x^2 - 2x = 0 \\\\\implies x(x-2) = 0 \\\\\implies\red{ x = 0 , 2 }

We got two values of x as 0 & 2 . Alternatively substituting these values we have ,

\implies y = 2 x - 3 \\\\\implies y = 2(0)-3 \qquad or \qquad y = 2(2)-3 \\\\\implies y = 0-3 \qquad or \qquad 4 - 3 \\\\\implies \red{ y = -3 , 1 }

Thefore the required answer is ,

\red{Option\:c} \begin{cases} (0,-3) \\ (2,1) \end{cases}

4 0
3 years ago
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