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dusya [7]
3 years ago
10

23. Which is a solution to the inequality below?

Mathematics
1 answer:
LenaWriter [7]3 years ago
8 0

Answer:

C.

Step-by-step explanation:

- 5n - 10 > 20

add 10 to both side

so...

- 5n > 30

divide by -5 so, the sign flips because multiplying or dividing with a negative number changes the sign.

30/5=6

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natka813 [3]
F(g(2))
g(2)=2^2+1=4+1=5
f(5)=5-3=2
f(g(2))=2
8 0
3 years ago
There are 86 people coming to my party and each person will need a cup how many packages of cups will i buy if one package of cu
marusya05 [52]

Answer:

i think its 4 packages

Step-by-step explanation:each package is 25 so if you get 2 its 50 + another package would be 75  so you need one more.

hope this helps

5 0
3 years ago
Read 2 more answers
Tools
tensa zangetsu [6.8K]

Answer:

6.7 to 26

Step-by-step explanation:

7 0
3 years ago
Can anyone answer this
notka56 [123]
I can’t seem to look at the bottom two answers and the two ( A and B) are not correct. So you’re left with either C or D
8 0
3 years ago
Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10
zvonat [6]

The approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

FV=P\left(1-\dfrac{r}{100}\right)^n

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is n_1. Thus, by the above formula for the first car,

0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}

Take log both the sides as,

\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is n_2. Thus, by the above formula for the second car,

0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}

Take log both the sides as,

\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14

The difference in the ages of the two cars is,

d=4.85-3.14\\d=1.71\rm years

Thus, the approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

brainly.com/question/25297296

4 0
2 years ago
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