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Varvara68 [4.7K]
3 years ago
14

Harry kept a weather journal. The sun was shining four out of five days. On what percent of the days was the sun shining?

Mathematics
2 answers:
Rufina [12.5K]3 years ago
8 0

Answer:

50

Step-by-step explanation:

sladkih [1.3K]3 years ago
7 0

Answer:

80%

Step-by-step explanation:

divide the number of days that there was sun (4) by the total number fo days (5), then multiply by 100 to get to a percent.

if you have any questions, leave them in the comments and i will try to answer them, if this helped, please give brainliest.

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Describe how you regroup when you find the sum of 64 + 43
Leno4ka [110]
All i do is i think of 4+3 which equals 7 and then 6+4 which equals 10 then i put the numbers together.

so my answer is 107

         64
         43
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        107
4 0
3 years ago
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This one too please!!!!!!!!!
Vinvika [58]
X= 0 and 1 are the answers
7 0
3 years ago
15 miles and 10 hours would be what miles per hour
suter [353]

Answer: 1.5miles/hour

Step-by-step explanation:

7 0
3 years ago
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Segments CD, AE, and BF are medians of triangle ABC.
EleoNora [17]

Option C: x = 5.5 is the value of x

Explanation:

Given that CD, AE, an BF are the medians of the triangle ABC

Also, given that AM = 4x - 5 and ME=x+3

We need to determine the value of x.

Since, we know that the centroid divides the median in the ratio 2 : 1

Hence, we have,

AM=2\times ME

Substituting the values, we get,

4x-5=2(x+3)

Simplifying, we get,

4x-5=2x+6

Subtracting both sides of the equation by 2x, we have,

2x-5=6

Adding both sides of the equation by 5, we have,

2x=11

Dividing both sides of the equation by 2, we get,

x=5.5

Therefore, the value of x is 5.5

Hence, Option C is the correct answer.

7 0
3 years ago
Consider the differential equation:
Wewaii [24]

(a) Take the Laplace transform of both sides:

2y''(t)+ty'(t)-2y(t)=14

\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s

where the transform of ty'(t) comes from

L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)

This yields the linear ODE,

-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s

Divides both sides by -s:

Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}

Find the integrating factor:

\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C

Multiply both sides of the ODE by e^{3\ln|s|-s^2}=s^3e^{-s^2}:

s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}

The left side condenses into the derivative of a product:

\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}

Integrate both sides and solve for Y(s):

s^3e^{-s^2}Y(s)=7e^{-s^2}+C

Y(s)=\dfrac{7+Ce^{s^2}}{s^3}

(b) Taking the inverse transform of both sides gives

y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that \frac{7t^2}2 is one solution to the original ODE.

y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7

Substitute these into the ODE to see everything checks out:

2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14

5 0
3 years ago
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