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sertanlavr [38]
3 years ago
14

PLEASE HELLPPPPPPPPPPPPPPPPPPPPPPPPPPP

Mathematics
1 answer:
suter [353]3 years ago
7 0

Answer:

a) 33

B) 65?

C) 60?

D) Becuase its the same shape just smaller.

E) 14, its cut in half

Step-by-step explanation:

be careful if you use these, its been a while since I geometry.

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What are the square roots of 64/144 ?
nordsb [41]

Answer: third option

Step-by-step explanation:

To solve the problem you must apply the proccedure shown below:

- Descompose the numerator and the denominator of the given fraction into its prime numbers:

64=2*2*2*2*2*2\\144=2*2*2*2*3*3

Then you can rewrite:

64=2^6\\144=2^4*3^2

Then:

\±\sqrt{\frac{{2^6}}{{2^4*3^2}}}=\±\frac{2^3}{2^2*3}=\±\frac{8}{4*3}=\±\frac{8}{12}

Therefore, the answer is: -\frac{8}{12}\ and\ \frac{8}{12}

4 0
4 years ago
Read 2 more answers
Find the exponential function P(t)=P0a^t where P(2)=5 and P(4)=10
asambeis [7]

Answer:

P(t)=\frac{5}{2} \cdot (\sqrt{2})^t

Step-by-step explanation:

P(2)=5 meants when t=2, that the value for P(t) is 5.

So this gives us this equation:

5=P_0 \cdot a^2

P(4)=10 meants when t=4, that the value for P(t) is 10.

So this gives us this equation:

10=P_0 \cdot a^4

So I take equation 2 and divide it be equation 1 I get:

\frac{10}{5}=\frac{P_0 \cdot a_4}{P_0 \cdot a_2}

Simplifying:

2=a^2

Since the base for an exponential function can't be negative then a=\sqrt{2}.

So plugging into one of my equations I began with gives me an equation to solve for the initial value,P_0:

5=P_0 \cdot (\sqrt{2})^2

5=P_0 \cdot 2

Divide both sides by 2:

\frac{5}{2}=P_0

The function is:

P(t)=\frac{5}{2} \cdot (\sqrt{2})^t

7 0
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Answer:multiplication Ex: 3 x 2 = 6

Division Ex: 6/2 =3

Step-by-step explanation:

3 x 2 = 3+3, add 3 2 times

6/2 = How many 2’s go into 6

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astra-53 [7]
The answer would be 5 I think
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Answer: The base area, diameter, and radius can be found out by discovering the volume and height. Substitute the values to the corresponding variables into the formula, and then solve by simplifying and undoing operations to isolate the variable to find the diameter which is double the radius.

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