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3 years ago

2. A 48-foot rope is cut into three lengths. The second length of rope is twice the first

Mathematics

1 answer:

BigorU [14]3 years ago

5 0

Answer:

48/3.2=15

Got 3.2 from Second length is twice the third so we divide by 2 + 1 = 3

0.2 is from 1/5 = 0.2

3 is the third length

30 is second length and 15 is the first.

Step-by-step explanation:

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sunlight can penetrate a certain Lake to the depth of fifteen meters the lake is five and a half times deeper than the depth of

gizmo_the_mogwai [7]

Fifteen times Fifteen point Five equals 82.2 so the answer would be 82.2 Meters

4 0

3 years ago

Use the rule y = -x + 7 to fill in the blank. If (-1, y) is on the graph, then y = ___

STALIN [3.7K]

Answer:

Step-by-step explanation:

x=-1 so -(-1)+7=y

1+7=y

8=y

8 0

3 years ago

6+2 divided by 15 (76 x 100)= ??

larisa [96]

Answer:

Step-by-step explanation:

We have to use PEMDAS

P is parentheses so we turn it to 6 + 2 divided by 15(7600)

E is for exponents but we don't have any

M is for multiplication and we have that so its 6 + 2 divided by 114000

D is for division so now its 6 + 1.754

A is for addition so 7.75400

5 0

3 years ago

Read 2 more answers

Problem PageQuestion The mass of a radioactive substance follows a continuous exponential decay model, with a decay rate paramet

asambeis [7]

Answer:

t = 1.51

Step-by-step explanation:

Exponential Model

The exponential model is often used to simulate the behavior of a magnitude that either grow or decay in proportion to the existing amount of that magnitude.

The model can be expressed as

$M=M_oe^{kt}$

In this case, Mo is the initial mass of the radioactive substance and k is a constant which value is positive if the mass is growing or negative if the mass is decaying.

The value of k is not precisely given in the question, we are assuming $k=-0.2$

The model is now

$M=M_oe^{-0.2t}$

We are required to compute the time it takes the mass to reach one-half of its initial value:

$\displaystyle \frac{M_o}{2}=M_oe^{-0.2t}$

Simplifying

$\displaystyle \frac{1}{2}=e^{-0.2t}$

Taking logarithms

$\displaystyle ln\frac{1}{2}=ln(e^{-0.2t})=-0.2t$

Solving for t

$\displaystyle t=-\frac{ln\frac{1}{2}}{0.2}=1.51$

6 0

3 years ago

What is the slope of this graph? −1/3 3 1/3 −3

FromTheMoon [43]

I'm assuming there is probably a graph that goes with this problem so here is all the info I can give you with the four choices: If it is -1/3, the line will go DOWN 1, over 3 (just count on the graph) If it is 3, the line will go UP 3, over 1 If it is 1/3, the line will go UP 1, over 3 And if it it -3, it will go DOWN 3, over 1 (If the slope is 3 or -3, it would be quite steep compared to a slope of 1/3 or -1/3) Hope that helps!

4 0

3 years ago