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Masja [62]
3 years ago
13

Negative three times a number plus 4 is no more than the number minus 8 find the number

Mathematics
1 answer:
elena-14-01-66 [18.8K]3 years ago
3 0

Answer:

x \geqslant 3

[3, infinity)

Step-by-step explanation:

-3n + 4 < n-8

get all of the n's to the same side

-3n + 4 < n-8

+3n + 4 < +3n-8

4 < 4n-8

send the number (-8) off to the other side

4 < 3n-8

+8< 3n +8

12 < 3n

isolate variable

12/3 < 3n/3

4 < n

so

x \geqslant 3

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Cumpute:<br><br> 11squared-(12-4)squared + 3 =
VikaD [51]

Answer:

The result is 60

Step-by-step explanation:

We have to use this expression showing BODMAS rule. According to the rule, we must first solve the brackets. In our expression, the term within the bracket is (12-4) squared = (8) squared = 64.

Then we need to perform the addition followed by subtraction.

The expression becomes:

11 squared - 64 + 3

= 121 - 64 + 3

= 124 - 64

= 60

So, the result of the expression is 60.

5 0
3 years ago
The manufacturer's suggested retail price (MSRP) for a particular car is $25,425, and it is expected to be
arsen [322]

Answer:

(a) The linear depreciation function of the car which gives the worth of the car y after a number of years, t is given as follows;

y = 25,425 - 2,739 × t

(b) The value of the car 7 years from now is $6,252

(c) $2,739 per year

Step-by-step explanation:

The manufacture's suggested retail price (MSRP) for the car = $25,425

The amount the car is expected to be worth in 5 years = $11,730

(a) The linear depreciation is given as follows;

Depreciation \ Per \ Year \ = \dfrac{Cost \ of \, Asset -  Salvage \ Value}{Life \ of \, Asset \ in \ use}

Where;

Cost of Asset = $25,425

Salvage Value = $11,730

Life of Asset in use = 5 years

We get;

Depreciation \ Per \ Year \ = \dfrac{\$ 25,425 -  \$11,730}{5} = \$2,739/year

Therefore, the linear depreciation function, can be written as follows;

y - 25,425 = -2,739×(t - 0)

y = -2,739·t + 0 + 25,425 = 25,425 -2,739·t

y = 25,425 - 2,739 × t

Where;

y = The expected worth of the car after a given number of years

t = The number of years used for the calculation of the depreciation

(b) The value of the car 7 years from now is given by substitution as follows;

Whet t = 7, we have;

y = 25,425 -2,739·t = y = 25,425 -2,739 × 7 = $6,252

The value of the car 7 years from now = $6,252

(c) Depreciation \ Per \ Year \ = \dfrac{\$ 25,425 -  \$11,730}{5} = \$2,739/year

The car is depreciating at a rate of $2,739 per year.

3 0
3 years ago
What is the scale factor?
Sever21 [200]

Answer:

I think the scale factor is 2 because 12 divided by 6 is 2.

8 0
3 years ago
Suppose your heart is beating 72 times per minute. At this rate, how many times will it beat in 15 minutes?
Vlad1618 [11]
Hey there!

All you need to do is multiply the beats per minute by 15 minutes.
72 x 15=1080. Your heart will beat 1080 times in 15 minutes.

I hope this helps!
8 0
4 years ago
What is 238% of 250????
Mariulka [41]

Answer:

595

Step-by-step explanation:

just convert 238% to a decimal (2.38) and multiply it by 250

4 0
3 years ago
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