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Ludmilka [50]
2 years ago
10

when the base of a ladder is placed 10ft away from the base of a building, the top of the ladder touches the building at 18ft ab

ove the ground. About how long is ladder?
Mathematics
1 answer:
Vikki [24]2 years ago
6 0
35 feet i think i’m not sure
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The equation is:
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4n^(2)+4n=3<br> Please help with the work
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Answer:

n= 1/2 ( \frac{1}{2} )  or  (0.5)

AND

n= -3/2 ( -\frac{3}{2} ) or (-1.5)

Step-by-step explanation:

4n^2+4n=3

step1: move 3 to the other side and make the equation equal to zero.

4n^2+4n-3=0

step2: factorise the equation.

(2n-1)(2n+3)=0

step3: make each bracket equal to zero.

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step4: solve for n values.

<u>1.</u> 2n-1=0

(add 1 for both sides)

2n=1

(divide by 2 for both sides)

n= 1/2 ( \frac{1}{2} )  or  (0.5)

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(divide by 2 for both sides)

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2 years ago
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}

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

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

464 grams.

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