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3 years ago

help with geometry please, i set up the equation like 5x+9=3x+11 and got 1 but that's not the right answer

Mathematics

1 answer:

Ugo [173]3 years ago

8 0

So, you put the values on the other sides of the vertical angles, so you get a set of interior angles (can't remeber word at the moment)
Make the angles equal to 180.
So equation looks like 3x+11+5x+9=180.
Now combine like terms, get 8x+20=180.
Subtract 20 from both sides and get 8x=160.
Divde both sides by 8 and get your answer of x= 20
Hope this helps

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satela [25.4K]

Answer:

$\large\boxed{2a+14}$

Step-by-step explanation:

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3 years ago

A recipe calls for 1/3 cup of grated Parmesan cheese for 8 servings of spaghetti sauce. How many servings can be made with 2 cup

Darya [45]

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4 years ago

Identify each coefficient <br> 1. 8x___

masha68 [24]

8 would be the coefficient :)

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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

Someone please help with this asap!! I will mark brainliest if it's right! Serious answers only !!!

Arlecino [84]

Answer:

1) given

2) given

3) definition of vertical angles

4) AAS

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers