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galina1969 [7]
3 years ago
10

The length of a school bus is 12.6 meters. If 9 school buses park end to end with 2 meters between each one, what's the total le

ngth from the front of the first bus to the end of the last bus?
Mathematics
1 answer:
weqwewe [10]3 years ago
7 0

Answer:

129.4

Step-by-step explanation:

Calculation for what's the total length from the front of the first bus to the end of the last

First step is to multiply the given number of buses by the length of each one

Hence,

12.6 meters x 9 = 113.4 meters

Second step

Since there are 8 spaces in between we would multiply the 8 spaces by 2 meters

8 x 2 meters = 16

Now let calculate total length from the front of the first bus to the end of the last

16 + 113.4 meters =129.4

Therefore total length from the front of the first bus to the end of the last will be 129.4

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Irene is N years old. If Tom is twice as old, which of the following algebraic statements shows h
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Answer:

1. 2n

Step-by-step explanation:

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3 years ago
If x+1/x = 5 find the value of x^3 +1/x^3. pls solve​
Korolek [52]

Answer:

1) solve x+1/x = 5

\frac{x + 1}{x}  = 5

x + 1 = 5x

x - 5x =  - 1

- 4x =  - 1

x =  \frac{ -1}{ - 4}

x =  \frac{1}{4}

2) solve x³+1/x³

x {}^{3}  +  \frac{1}{x {}^{3} }

substitute x = 1/4 into the expression

( \frac{1}{4} ) {}^{3}  +  \frac{1}{( \frac{1}{4}) {}^{3}  }

\frac{1}{64}  +  \frac{1}{ \frac{1}{64} }

\frac{1}{64}  +  \frac{64 \times 1}{1}

\frac{1}{64}   + 64

\frac{4097}{64}  \: or \: 64.02

5 0
3 years ago
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Sati [7]
What do you need help with?
5 0
3 years ago
Someone help me solve 29*26/54.33a^26b-893d
azamat

Answer:

The answer is 25g^R4

Step-by-step explanation:

When you solve the equation, it becomes 25g^R4, it doesn't result in a whole number.

6 0
3 years ago
Read 2 more answers
Question 3 (1 point) Co-60 has a half life of 5.3 years. If a pellet that has been in storage for 26.5 years contains 14.5g of C
Zarrin [17]

Answer:

464 grams.

Step-by-step explanation:

Amount of substance:

The amount of substance after t years is given by an equation in the following format:

A(t) = A(0)(1-r)^t

In which A(0) is the initial amount and r is the decay rate, as a decimal.

Co-60 has a half life of 5.3 years.

This means that:

A(5.3) = 0.5A(0)

We use this to find r. So

A(t) = A(0)(1-r)^t

0.5A(0) = A(0)(1-r)^{5.3}

(1-r)^{5.3} = 0.5

\sqrt[5.3]{(1-r)^{5.3}} = \sqrt[5.3]{0.5}

1 - r = 0.5^{\frac{1}{5.3}}

1 - r = 0.8774

So

A(t) = A(0)(0.8774)^{t}

If a pellet that has been in storage for 26.5 years contains 14.5g of Co-60, how much of this radioisotope was present when the pellet was put in storage?

We have that A(26.5) = 14.5, and use this to find A(0). So

A(t) = A(0)(0.8774)^{t}

14.5 = A(0)(0.8774)^{26.5}

A(0) = \frac{14.5}{(0.8774)^{26.5}}

[tex]A(0) = 464[tex]

464 grams.

3 0
3 years ago
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