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vaieri [72.5K]
3 years ago
7

Bruhhh i need help asap

Mathematics
1 answer:
Nataly_w [17]3 years ago
4 0
With what??kfjyjyjjtjhjhjkroeowoeieididi
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Maria studied the traffic trends in India. She found that the number of cars on the roads increases by 10% each year. If there w
laiz [17]
You would do

(1.1)(80,000,000)=88,000,000

That is the total cars for year two. To find the increase you would do

(1.1)(88,000,000)=96,800,000

Then you would subtract year 2 from year three

96,800,000-88,000,00=8,800,000

That gives you ur answer.
7 0
3 years ago
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Plsss help will mark brainliest
DiKsa [7]
C, because 7 divided, or over nine, is 7/9
Hope this helps!
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3 years ago
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3.3 = ? as a friction
Allushta [10]

Answer:

33/10

Step-by-step explanation:

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3 years ago
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Prove the divisibility:<br><br>45^45·15^15 by 75^30
garri49 [273]

Answer:

3^{75}.

Step-by-step explanation:

We have been an division problem: \frac{45^{45}*15^{15}}{75^{30}}.

We will simplify our division problem using rules of exponents.

Using product rule of exponents (a*b)^n=a^n*b^n we can write:

45^{45}=(9*5)^{45}=9^{45}*5^{45}

15^{15}=(3*5)^{15}=3^{15}*5^{15}

75^{30}=(15*5)^{30}=15^{30}*5^{30}

Substituting these values in our division problem we will get,

\frac{9^{45}*5^{45}*3^{15}*5^{15}}{15^{30}*5^{30}}

Using power rule of exponents a^n*a^m=a^{n+m} we will get,

\frac{9^{45}*5^{(45+15)}*3^{15}}{15^{30}*5^{30}}

\frac{9^{45}*5^{60}*3^{15}}{15^{30}*5^{30}}

Using product rule of exponents (a*b)^n=a^n*b^n we will get,

\frac{(3*3)^{45}*5^{60}*3^{15}}{(3*5)^{30}*5^{30}}

\frac{3^{45}*3^{45}*5^{60}*3^{15}}{3^{30}*5^{30}*5^{30}}

Using power rule of exponents a^n*a^m=a^{n+m} we will get,

\frac{3^{(45+45+15)}*5^{60}}{3^{30}*5^{(30+30)}}

\frac{3^{105}*5^{60}}{3^{30}*5^{60}}

\frac{3^{105}}{3^{30}}

Using quotient rule of exponent \frac{a^m}{a^n}=a^{m-n} we will get,

\frac{3^{105}}{3^{30}}=3^{105-30}

3^{105-30}=3^{75}

Therefore, our resulting quotient will be 3^{75}.

7 0
3 years ago
Find the missing angles 1-3
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Answer:

ssdfggbbdbdbdbdbd d sbsbbsnsndnndnddhd

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