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JulijaS [17]
3 years ago
8

Determine whether 96 is divisible by 3

Mathematics
2 answers:
aleksandr82 [10.1K]3 years ago
7 0
96 divided by 3 is 32
Therefor, it is divisible by 3
Is this what you were looking for?
Bas_tet [7]3 years ago
3 0
Yes it is divisible by 3.

When doing long division, you get the answer 32 with no remaining.

Since there is no remaining, its a divisible by 3.

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Vlad1618 [11]

Hello from MrBillDoesMath!

Answer:

35x - 600

Discussion:

Area of unshaed rectangle =

area of full rectangle - area of shaded rectangle =

35x - 600

Thank you,

MrB

3 0
3 years ago
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Which polynomial has factors of 4x – 7 and x + 4?
34kurt
B. because when you multiply the factors they will get B
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Which of the following is an example of exponential growth or decay?
Aleksandr [31]
I think the answer is B)
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Ma Bernier has 52 cats and dogs in her house the number of cats is three times the number of dogs how many cats and dogs are in
frosja888 [35]

Answer:

dogs - 13

cats - 39

Step-by-step explanation:

Let the number of dogs Ma Bernier have be represented with d

She has 3 times as many cats as dogs, the number of cats she has :

cats = 3d

The sum of the cats and dogs is 52

d + 3d = 52

4d = 52

divide both sides of the equation by 4

d = 13

She has 13 dogs

Cats = 3d

= 3 x 13 = 39

3 0
3 years ago
Pls help asap whats the local min value of the function below?
shutvik [7]

Answer:

<em>given function has 2 minimums</em> - \frac{9}{4}  and  \frac{9}{4}  

Step-by-step explanation:

<u><em>Step 1.</em></u> g'(x) = 4x³ - 10x

<u><em>Step 2.</em></u> Find find the critical points:

4x³ - 10x = 2x(2x² - 5) = 0

x_{1} = - \sqrt{\frac{5}{2} } , x_{2} = 0 , x_{3} = \sqrt{\frac{5}{2} }

<u><em>Step 3.</em></u> g'(x) > 0 :  - \sqrt{\frac{5}{2} } < x < 0  or  x > \sqrt{\frac{5}{2} }

g'(x) < 0 :   x < - \sqrt{\frac{5}{2} }   or   0 < x < \sqrt{\frac{5}{2} }

<u><em>Step 4.</em></u>

If x ∈ ( - ∞ , - \sqrt{\frac{5}{2} } ) , g(x) is decreasing ;

If x = - \sqrt{\frac{5}{2} } , g(x) has <em>minimum</em> value ;

If x ∈ ( - \sqrt{\frac{5}{2} } , 0 ) , g(x) is increasing ;

If x = 0 , g(x) has maximum value ;

If x ∈ ( 0 , \sqrt{\frac{5}{2} } ) , g(x) is decreasing ;

If x = \sqrt{\frac{5}{2} } , g(x) has <em>minimum</em> value ;

If x ∈ ( \sqrt{\frac{5}{2} } , ∞ ) , g(x) is increasing .

⇒ at ( - \sqrt{\frac{5}{2} } , - \frac{9}{4} ) and at ( \sqrt{\frac{5}{2} } , \frac{9}{4} ) , g(x) reaches its minimum

8 0
3 years ago
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