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elixir [45]
3 years ago
6

A 222-inch pipe is cut into two pieces. One piece is five times the length of the other. Find the length of the shorter piece.

Mathematics
1 answer:
viktelen [127]3 years ago
5 0

Answer:

The shorter piece is <u>37</u> inches long.

Step-by-step explanation:

Let the length of the shorter piece be x and the longer piece be 5x.

Short piece + Long piece = 222 inches

x + 5x = 222

6x = 222

x = 222 ÷ 6

x = 37

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14x^2-17x-6. (^being 14x squared)
zysi [14]
Factor?, if so
if in ax^2+bx+c form then
multiply a and c together
result=z,
find 2 numbers that multiply to get z and add up to get b
 so



14=a
-17=b
-6=c
14 times -6=-84
factor to find what 2 numbers add up to -17
-84=
-2, 42
-3,28
-4,21
-6,14
-7,12

add each up

-2, 42=40 nope
-3,28=25 nope
-4,21=17 yes, becase if -21 and 4 then answer
-6,14=8 nope
-7,12=5 nope

split up -17 into -21 and 4


14x^2-21x+4x-6
group
(14x^2-21x)+(4x-6)
undistribute
(7x)(2x-3)+(2)(2x-3)
revesres undistribute (ab+ac=a(b+c) or (a+b)(c)+(a+b)(d)=(a+b)(c+d))
(7x+2)(3x-3) is factore out form

then set it to zero and do th emath because if xy=0 then assume x and/or y=0 so
7x+2=0
subtract 2
7x=-2
divide by 7
x=-2/7

3x-3=0
add 3
3x=3
divide 3
x=1
x=-2/7 or 1

6 0
4 years ago
The half-life of a certain radioactive substance is 45 days. There are 6.2 grams present initially. On what day
kvasek [131]

Answer:

There will be less than 1 gram of the radioactive substance remaining by the elapsing of 118 days

Step-by-step explanation:

The given parameters are;

The half life of the radioactive substance = 45 days

The mass of the substance initially present = 6.2 grams

The expression for evaluating the half life is given as follows;

N(t) = N_0 \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{1/2}}

Where;

N(t) = The amount of the substance left after a given time period = 1 gram

N₀ = The initial amount of the radioactive substance = 6.2 grams

t_{1/2} = The half life of the radioactive substance = 45 days

Substituting the values gives;

1 = 6.2 \left (\dfrac{1}{2} \right )^{\dfrac{t}{45}

\dfrac{1}{6.2}  =  \left (\dfrac{1}{2} \right )^{\dfrac{t}{45}

ln\left (\dfrac{1}{6.2} \right )  =  {\dfrac{t}{45} \times ln \left (\dfrac{1}{2} \right )

t = 45 \times \dfrac{ln\left (\dfrac{1}{6.2} \right ) }{ln \left (\dfrac{1}{2} \right )} \approx 118.45 \ days

The time that it takes for the mass of the radioactive substance to remain 1 g ≈ 118.45 days

Therefore, there will be less than 1 gram of the radioactive substance remaining by the elapsing of 118 days.

3 0
3 years ago
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kvasek [131]

Answer:

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Step-by-step explanation:

3+5=8

8 to the 3rd power= 512

512 times 5= 2560

2560+1= 2561

8 0
3 years ago
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It's (5,2) , I took it already so here you go
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Answer:

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Step-by-step explanation:

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3 years ago
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