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3 years ago

PLEASE HELP! BRAINLIEST!! ANSWER ASAP!!!

Mathematics

2 answers:

spayn [35]3 years ago

5 0

Enrollment at X = 0 is at 20.

Enrollment at X = 8 is 120.

Find the difference between those values:

120 - 20 = 100

Divide that difference by the difference in the x values:

100 /8 = 12.5

The answer is D. 12.5

never [62]3 years ago

3 0

Answer it will be D yes

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Two angles of a triangle are the same size. The third angle is 3 times as large

Gnoma [55]

Use x to represent the first two angles. If the third angle is 3 times as large as the first, it can be represented as 3x. So, your angles are x, x, and 3x. These need to all add to 180 degrees, since it’s a triangle.

x + x + 3x = 180 —> Simplify the left side.
5x = 180 —> Divide by 5.
x = 36

So the first two angles are 36 degrees, and the third angle is 3 times that, which is 108 degrees.

3 0

3 years ago

What is the measure of the base in the triangle shown below?<br> A. 11<br> B. 45<br> C. 25<br> D. 30

dangina [55]

Answer:

c. 25

Step-by-step explanation:

BOTH SIDES HAVE X SO YOU PUT THEM IN A 1 EQUATION WAY AND SOLVE X:

4X+1=2X+23

4X-2X=23-1

2X=32

X= 32/2=11

THEN YOU GO TO THE BASE AND CHANGE X AND SOLVE THE EQUATION WITH X=11

3x11-8

33-8= 25=BASE

5 0

3 years ago

A police officer estimated the distance from one side of a street to the other to be 14m . The actual distance was 13.4.

Oksi-84 [34.3K]

The absolute error is 0.2 and the percent error is 1.45 I hope this helps

7 0

3 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

A bucket full of milk is 80cl,how many litres are in 15 of such

PIT_PIT [208]

Answer:

1.2 liters

Step-by-step explanation:

1 bucket has a volume of 80 cL.

15 buckets have a volume of 15 * 80 cL = 120 cL

1 L = 100 cL

120 cL * (1 L)/(100 cL) = 1.2 L

5 0

3 years ago