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egoroff_w [7]
3 years ago
7

A school had 225 students enrolled last year. This year, 260 students are enrolled. Find the percent of change to the

Mathematics
1 answer:
Katarina [22]3 years ago
8 0

Answer:

15.5%

Step-by-step explanation:

No of students that had enrolled last year = 225

This year, 260 students are enrolled.

We need to find the percent of change in the number of students. It can be calculated as :

P = (change in no of students/initial no of students)×100

\%=\dfrac{260-225}{225}\times 100\\\\=15.5\%

So, the percent change is 15.5%.

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In ΔHIJ, the measure of ∠J=90°, JI = 5, IH = 13, and HJ = 12. What is the value of the sine of ∠H to the nearest hundredth?
Arte-miy333 [17]

<u>Given</u>:

Given that HIJ is a right triangle.

The measure of ∠J is 90°, JI = 5, IH = 13, and HJ = 12.

We need to determine the value of sine of ∠H

<u>Value of sine ∠H:</u>

The value of sine ∠H can be determined by using the trigonometric ratios.

Thus, we have;

sin \ H=\frac{JI}{IH}

Substituting the values, we get;

sin \ H=\frac{5}{13}

Dividing, we get;

sin \ H=0.3846

Taking sin^{-1} on both sides, we have;

H=sin^{-1}(0.3846)

H=22.6189

Rounding off to the nearest hundredth, we get;

H=22.62^{\circ}

Thus, the measure of ∠H is 22.62°

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3 years ago
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Y= -3x - 1 for x= 5????
adell [148]

Answer:

y= -16

Step-by-step explanation:

Y= -3x - 1

Let x = 5

y = -3(5) -1

y = -15-1

y = -16

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3 years ago
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On a test spencer got 1/8 of the problems worng . which set best represents the ones he got worng
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H(x)=3x^2−12x+16 what does h equal
Verizon [17]

Answer:

(x)=3x^2−12x+16

Answer- =3x2−12x+16

Step-by-step explanation:

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Which of the following integrals represents the area of the region bounded in the first quadrant by x = pi/ 4 and the functions
Kaylis [27]

Answer:

option B is true.

Step-by-step explanation:

We are given that two functions

f(x)=sec^2x and g(x)=sin x and a line x =\frac{\pi}{4}

We have to find the area of the region bounded in the first quadrant by x={\pi}{4} and two functions

We know that the area bounded by two functions

=Integration of region(Upper curve- lower curve)

Therefore, function of sec square x is upper curve and function of sin x is lower function

Therefore, limit of x changing from 0 to \frac{\pi}{4}

Hence, the area of the region bounded in the first quadrant and two functions is given by

=\int_{0}^{\frac{\pi}{4}} (sec^2x-sinx) dx

Therefore, option B is true.

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