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KIM [24]
3 years ago
14

Divide 28 in the ratio of 3:1

Mathematics
2 answers:
PolarNik [594]3 years ago
7 0
The answer to this is 28:3
Amiraneli [1.4K]3 years ago
6 0
Add up 3 and 1:  3+1 = 4.  The corresponding fractions are 3/4 and 1/4.

(3/4)(28) =21
(1/4)(28)  = 7

Then 21 and 7 add up to 28, as required, and are in the ratio 3:1.
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on monday, roger drove to work in 20 minutes. on tuesday, he averaged 15 miles per hour more, and it took him 6 minutes less to
ch4aika [34]
Let the distance to his office be x, then
On monday, speed = x/20 miles per minutes = 3x miles per hour
On tuesday, speed = (3x + 15) miles per hour
Time = distance / speed
(20 - 6)/60 hours = x/(3x + 15) hours
7/30 = x/(3x + 15)
7(3x + 15) = 30x
21x + 105 = 30x
30x - 21x = 105
9x = 105
x = 105/9 = 11.7 miles

Therfore he travels an average of 11.7 miles to work.
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3 years ago
Find the y-intercept of the line on the graph
hoa [83]
2 it’s literally just where the line crosses on the y axis
8 0
2 years ago
The zeros of function dare -3 and 8. Which expression could be a factor of function d?
erastovalidia [21]

Answer:

(x + 3)

Step-by-step explanation:

Using the zero product property

x-a = 0  x-b = 0  where a and b are the zeros

(x-a)(x-b) =0

(x- -3)(x -8) =0

(x+3) (x-8) =0

3 0
2 years ago
What is the decimal equivalent of 1/3
Vesna [10]

Answer:

Step-by-step explanation:

1/3 0.333 repeated 0.333 times 100 =33.333…%

4 0
2 years ago
Read 2 more answers
If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,
julsineya [31]
<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

<h2>Why?</h2>

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)

(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0
3 years ago
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