Distribute: -3(4a-5b+c)

Mathematics

1 answer:

kvv77 [185]2 years ago

4 0

Answer:

I got it.

Step-by-step explanation:

-12a - (-15b) + (-3c)

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A restaurant manager wants to purchase 200 sets of dishes. one design costs $15 per set, while another costs $55 per set. if

vivado [14]

Hello,

Let's solve this problem mathematically.

For first design...

$f = 200 \times 15 = 3000$

For second design...

$s = 200 \times 55 = 11000$

To use an entire budget for design she must order.

For first design

5800÷3000 = 1.93f

or
For second design

5800÷11000=0.52s

Hope this helps answer your question!
r3t40

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2 years ago

What is the value of z? Plz help

Vera_Pavlovna [14]

Answer: 24

Step-by-step explanation: 58 = 2z - 10

58 - 10 = 48 = 2z

z = 24

6 0

2 years ago

Mr.Driscoll has 9 reports to grade.There are 6 pages for each report.He grades 12 pages.How many pages does he still have to gra

Juli2301 [7.4K]

42 pages left to grade and 7 reports left

8 0

2 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

2 years ago

Write the equation for the inverse of y=4arccot(

garri49 [273]

I think you meant "y = 4 arccot (x)."

This is equivalent to:

y
---- = arccot x
4

and so cot (y/4) = x

8 0

2 years ago

Distribute: -3(4a-5b+c)​

Distribute: -3(4a-5b+c)