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svetlana [45]
3 years ago
14

What is 3/10x9 please help me

Mathematics
2 answers:
Monica [59]3 years ago
6 0
27/10 is the answer
lora16 [44]3 years ago
5 0
(3/10)×9

=27/10=2.7  
>>>>>>>>>>>>>
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Help me please!!!!!!!!!!!!
iVinArrow [24]

Answer: 0.25g<2.50.... g<10

Step-by-step explanation: Let us say that the number of gumballs bought is represented by the variable g. In this case, the question is asking how many gumballs can be bought without surpassing the price of $2.50. We know that each gumball is $0.25, therefore the number of gumballs we buy times $0.25 has to be less than $2.50. Hence, the inequality would be 0.25g<2.50. If we were to solve this then g<2.50/0.25-----> g<10. In conclusion, the number of gumballs you can buy has to be less than 10. Thank you!

6 0
3 years ago
What values of b satisfy 3(2b + 3)2 = 36?
KiRa [710]

the correct question is

What values of b satisfy 3(2b+3)^2 = 36


we have

3(2b+3)^2 = 36

Divide both sides by 3

(2b+3)^2 = 12

take the square root of both sides

( 2b+3)} =(+ /-) \sqrt{12} \\ 2b=(+ /-) \sqrt{12}-3


b1=\frac{\sqrt{12}}{2} -\frac{3}{2}

b1=\sqrt{3} -\frac{3}{2}


b2=\frac{-\sqrt{12}}{2} -\frac{3}{2}

b2=-\sqrt{3} -\frac{3}{2}

therefore


the answer is

the values of b are

b1=\sqrt{3} -\frac{3}{2}

b2=-\sqrt{3} -\frac{3}{2}


6 0
3 years ago
Read 2 more answers
2 PUIHS
tatiyna

Answer:

45%

Step-by-step explanation:

3 0
3 years ago
Help I need to complete this before 3:00 pm
Oduvanchick [21]
It would be 64 x 8 x 8 =
3 0
3 years ago
If f(x) = x2 - 1 and g(x) = 2x - 3, what is the domain of (fog)(x)?
ladessa [460]

Answer:

domain will be ( -∞, ∞)

Step-by-step explanation:

The given functions are f(x) = x² - 1 and g(x) = 2x - 3

We have to find domain of (fog)(x)

We will find the function (fg)(x) first.

(fog)(x) = f[g(x)]

         = (2x - 3)²

         = 4x² + 9 - 12x - 1

        = 4x² - 12x + 8

       = 4 (x² - 3x + 2)

The given function is defined for all values of x.

Therefore, domain will be ( -∞, ∞)

<u>brainly.com/question/2458431</u>

4 0
2 years ago
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