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4vir4ik [10]
2 years ago
5

The student enrollment of South High School was 1,350 students in 1995 and increased by 15% per year until 2010. Approximately h

ow many students were enrolled at South High School in 2010?
A.About 10,985 students
B.About 9,552 students
C.About 5,366 students
D.About 5,740 students
Mathematics
2 answers:
vlabodo [156]2 years ago
5 0

Answer: A.

Step-by-step explanation:

Lerok [7]2 years ago
5 0

Answer:

A (10,985)

Step-by-step explanation:

y= 1,350(1+0.15)^15

1,350 is the student =a

0.15 which originally was 15% is the rate. =r

^15 is the time because 1995 to 2010 took 15 years. =t

The equation I use for this question:

y=a(1+r)^t  a.k.a Exponential growth formula

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Which rate is faster:a bird traveling at a constant rate of 2 kilometer per hour or an insect travelinf at a constant rate of 1,
Alla [95]
The bird since 1 kilometer is 1000 meters
6 0
3 years ago
Find the solution of 64 = 16u algebraically.<br> please I’m desperate
Sauron [17]

Answer:

u = 4

Step-by-step explanation:

Divide by 16 on both sides. [Division Property of Equality]

64/16 = 16u/16

u = 4

==================================================================<em>Hope I Helped, Feel free to ask any questions to clarify :) </em>

<em> Have a great day! </em>

<em> More Love, More Peace, Less Hate. </em>

<em>      -Aadi x</em>

3 0
2 years ago
Consider the equation below -2(5x + 8) = 14 + 6x complete the statements below with the process used to achieve steps 1-4
Ad libitum [116K]

Step-by-step explanation:

1. Distribute the -2. multiply-2 with 5x and 8. it will be -10x-16=14+6x. For me, i make small numbers negative or positive than big ones. Add 16 to both sides. it will be -10x=30+6x. Subtract 6x. -10x-6x=-16x. Divide both sides by 16. x=30/16= 1 and 14/16

6 0
2 years ago
Plz help me 8+x^2-11
likoan [24]

Answer:

-3 + x^2

Step-by-step explanation:

8+x^2-11

First combine the like terms.

so..

8-11 = -3

= -3 + x^2

but we don't know the value of x so we just leave it as it is.

And they both are not like terms so we cant solve them together so you stop there

Hope that helped!

6 0
2 years ago
Read 2 more answers
A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks along the x-axis from the spotlight toward the bui
Sedbober [7]

Answer:

0.675 m/s

Step-by-step explanation:

Let height of shadow= y,CD=x

Height of man=2 m

Speed of man= \frac{dx}{dt}=1. 8 m/s

\triangle ABD\sim\triangle ECD

Therefore, \frac{AB}{EC}=\frac{BD}{CD}

\frac{y}{2}=\frac{12}{x}

xy=24

Differentiate w.r.t t

x\frac{dy}{dt}+y\frac{dx}{dt}=0

x\frac{dy}{dt}=-y\frac{dx}{dt}

\frac{dy}{dt}=-\frac{y}{x}\frac{dx}{dt}

When the man is 4 m from  the building

Then, we have x=12-4=8 m

\frac{dx}{dt}=1.8 m/s

Substitute the values in above equation then, we get

8y=24

y=\frac{24}{8}=3

Substitute the values then we get

\frac{dy}{dt}=-\frac{3}{8}\times 1.8=-0.675 m/s

Hence, the length of his shadow on the building decreasing at the rate 0.675 m/s.

8 0
3 years ago
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