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Alex Ar [27]
2 years ago
7

What is 5x5????? I'll give brainliest :))))

Mathematics
2 answers:
Elan Coil [88]2 years ago
8 0

Answer:

the answer to 5×5 is 25

Step-by-step explanation:

your welcome

Mekhanik [1.2K]2 years ago
7 0

Answer:25

Step-by-step explanation:

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tensa zangetsu [6.8K]

Answer:

80, every 5 percent is 4 people. So every 25% is 20. 20 x 4 is 80.

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
What negative fraction equal 5/8
Mademuasel [1]
-10/-16 could be an answer or multiply the same number up and down but the number must be negative
6 0
3 years ago
Read 2 more answers
What is the value of this expression when a = 2 and b = -3 ? a^3 - b^3 / 5
saveliy_v [14]

Answer:

13 2/5

Step-by-step explanation:

a = 2 and b = -3

so the question asks whats.... a^3 - b^3/5

First we plug in the values of a and b

(2)^3 - (-3)^3 /5

Now we solve the ones in paranthesis first

(2)^3 = 8 because 2×2×2 and

-(-3)^3 forget about the - outside the parenthesis so

(-3)^3 = (-27) because (-3)×(-3)×(-3)

now we put it back together

8 -(-27)/5

the two minus become plus so

8 + 27/5

Now we solve it like fractions

8 and 27/5

simplify

13 and 2/5

Hope that helps!

5 0
3 years ago
5) Which expression is equivalent to 4(3 + 2)?
kirill [66]
D, because if you distribute 4(3+2) it equals= 4x3
= 12
4x2
=8
and there’s an addition sign so you will add the remaining numbers 12+8
3 0
2 years ago
A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns,
guajiro [1.7K]

We have been given that a huge ice glacier in the Himalayas initially covered an area of 45 square kilo-meters. The relationship between A, the area of the glacier in square kilo-meters, and t, the number of years the glacier has been melting, is modeled by the equation A=45e^{-0.05t}.

To find the time it will take for the area of the glacier to decrease to 15 square kilo-meters, we will equate A=15 and solve for t as:

15=45e^{-0.05t}

\frac{15}{45}=\frac{45e^{-0.05t}}{45}

\frac{1}{3}=e^{-0.05t}

Now we will switch sides:

e^{-0.05t}=\frac{1}{3}

Let us take natural log on both sides of equation.

\text{ln}(e^{-0.05t})=\text{ln}(\frac{1}{3})

Using natural log property \text{ln}(a^b)=b\cdot \text{ln}(a), we will get:

-0.05t\cdot \text{ln}(e)=\text{ln}(\frac{1}{3})

-0.05t\cdot (1)=\text{ln}(\frac{1}{3})

-0.05t=\text{ln}(\frac{1}{3})

t=\frac{\text{ln}(\frac{1}{3})}{-0.05}

t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-0.05\cdot 100}

t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-5}

t=-\text{ln}(\frac{1}{3})\cdot 20

t=-(\text{ln}(1)-\text{ln}(3))\cdot 20

t=-(0-\text{ln}(3))\cdot 20

t=20\text{ln}(3)

Therefore, it will take 20\text{ln}(3) years for area of the glacier to decrease to 15 square kilo-meters.

6 0
3 years ago
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