A scale factor of 4 is:<br> an enlargement<br> a reduction

Solve and check the following equation algebraically. 1/2(x + 32) = -10

ANTONII [103]

Solve :
1/2(x + 32) = -10....multiply both sides by 2
x + 32 = -10 * 2
x + 32 = -20
x = -20 - 32
x = - 52 <===

check :
1/2(x + 32) = -10.....when x = -52
1/2(-52 + 32) = -10
1/2(-20) = -10
-20/2 = -10
-10 = -10 (correct)

8 0

3 years ago

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Amanda's jewelry box is in the shape of a cube that has 6- inch edges what is the volume of Amanda's jewelry box

Dmitry [639]

So base on the question that describe the Amanda's jewelry box that is in shape of a cube and has a measurement of 6 inch on on its edge, to calculate the volume of the said cube, you must multiply the said number to the power of 3 and with that, the volume would be 216 cubic inches

7 0

3 years ago

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Please help! Will get brainliest!!!

ASHA 777 [7]

Chart;the answer is b

and it makes sense answer is ; also b

3 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

Tell whether the system has one solution, infinitely many solutions, or no solution.

nadya68 [22]

X=-7y+34
x+7y=32

To begin with make the two problems the same kind of problem. By adding 7y on both sides for the first problem.
x+7y=34
x+7y=32

So the correct answer would be no solution because it's the same problem and it can't equal to different answers.

8 0

2 years ago

A scale factor of 4 is: an enlargement a reduction