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2 years ago

I need your help guys please

Mathematics

1 answer:

myrzilka [38]2 years ago

4 0

1) 2x+14=x+18
2x=x+4
x=4

2) 9y-2=3y+10
9y=3y+12
6y=12
y=2

3) 2*(2[4]+14) + (9[2]-2)

4) 32/2=16

5) 11*2=22

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PLEASE HELP WILL GIVE BRAINLIEST

quester [9]

Answer:

The answer is the same for all of the boxes. It is 3 gallons.

Step-by-step explanation:

8 0

3 years ago

A bag contains 100 apples, 100 bananas, 100 oranges, and 100 pears. If I pick one piece of fruit out of the bag every minute, ho

k0ka [10]

Answer:

required time is here 45 minutes

Step-by-step explanation:

given data

apples = 100

bananas = 100

oranges = 100

pears = 100

solution

we have here 4 different kind of fruit

so that it is possible for each kind of fruit is

each kind of fruit is = 4 × 11 = 44

so each kind contain 11 fruit

as that 45th fruit make a dozen

now 1 fruit is pick every minute

'so we take here 45 minute to pick up 45 fruit

so that required time is here 45 minutes

5 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

What is 2,345 rounded to the nearest 10

ser-zykov [4K]

2,350 Hope this helped

8 0

3 years ago

Read 2 more answers

The mean of the values in a data set is c. If each of the values in the data set were multiplied by 6, what would be the mean of

oksian1 [2.3K]

Answer:

Mean of the data would be 6c

Step-by-step explanation:

<u>Rule to Remember:</u> Whenever each term of the data is multiplied by the same number, the new mean is previous mean multiplied by that number.

In this case, original value of mean was c. Each value of the data is multiplied with 6 so a result the new mean would be 6 times the original mean i.e. 6c

Let us consider that original values in the data set are x, y and z. Their mean would be:

$Mean=\frac{x+y+z}{3}$

If each value of the data is multiplied with 6, the new values would be 6x, 6y and 6z. The mean in this case will be:

$Mean=\frac{6x+6y+6z}{3}\\Mean=\frac{6(x+y+z)}{3}\\Mean=6\times\frac{x+y+z}{3}\\Mean=6c$

We can see from above that the new mean is 6 multiplied to the original mean.

Therefore, the answer to this question is 6c.

7 0

3 years ago

I need your help guys please ​

I need your help guys please