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2 years ago

If Deanna starts running at 7:30 and she runs for an hour and a half. How long does she end up running?

2 answers:

6 0

Deanna would be running up until 9:00 o'clock, this is because 7:30 plus I hr and a half equal to 8:60, and then that would turn into 9:00.

emmainna [20.7K]2 years ago

4 0

Hello!

Deanna starts running at 7:30, and runs for an hour and a half. Meaning, 1:30

To find out how long she ends up running, we must add the hour and a half to the time she started running:

7:30 + 1:30 = 9:00

Deanna ends up running until 9:00.

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360 kilometers is 24% of x kilometers its for the 14 points

ch4aika [34]

360=24%0f x
360=24/100*x
360*100=24x
36000/24=x
therefore, x=1500......

4 0

2 years ago

Read 2 more answers

The point nearest to the origin on a line is at (4, -4). Find the standard form of the equation of the line.

Scrat [10]

Just did a specific one of these; let's do the general case.

The point nearest the origin is (a,b).

The line from the origin through the point is

$bx - ay = 0$

The line we seek is perpendicular to this one. We swap the coefficients on x and y, negating one, to get the perpendicular family of lines. We set the constant by plugging in the point (a,b):

$ax + by = a^2 + b^2$

$ax + by -( a^2 + b^2) = 0$

That's standard form; let's plug in the numbers:

$4 x - 4 y - 32 = 0$

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5 0

3 years ago

Suppose that a large pumpkin empty a swimming pool in 40 hours and that a small pumpkin empty a pool and 70 hours working togeth

Norma-Jean [14]

Answer:

Total 25.45 hours will be taken to empty the pool.

Step-by-step explanation:

Time taken by large pumpkin to empty the pool=40 hours

Time taken by the smaller pumpkins to empty the pool=70 hours

Amount of pool emptied by large pumpkin in one hour= $\frac{1}{40}$

Amount of pool emptied by small pumpkin in one hour= $\frac{1}{70}$

Amount of pool emptied in one hour if both pumpkins work together=

$\frac{1}{40}$ + $\frac{1}{70}$

= $\frac{11}{280}$

Therefore, the no. of days will be the reciprocal of the amount of work done in one hour ie . $\frac{280}{11}$

Total 25.45 hours will be taken to empty the pool.

5 0

3 years ago

4x^2 y+8xy'+y=x, y(1)= 9, y'(1)=25

jarptica [38.1K]

Answer with explanation:

$\rightarrow 4x^2y+8x y'+y=x\\\\\rightarrow 8xy'+y(1+4x^2)=x\\\\\rightarrow y'+y\times\frac{1+4x^2}{8x}=\frac{1}{8}$

--------------------------------------------------------Dividing both sides by 8 x

This Integration is of the form ⇒y'+p y=q,which is Linear differential equation.

Integrating Factor

$=e^{\int \frac{1+4x^2}{8x} dx}\\\\e^{\log x^{\frac{1}{8}+\frac{x^2}{2}}\\\\=x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}$

Multiplying both sides by Integrating Factor

$x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\times [y'+y\times\frac{1+4x^2}{8x}]=\frac{1}{8}\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\\\\ \text{Integrating both sides}\\\\y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\frac{1}{8}\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx \\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=-[x^{\frac{9}{8}}]\times\frac{ \Gamma(0.5625, -x^2)}{(-x^2)^{\frac{9}{16}}}\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=(-1)^{\frac{-1}{8}}[ \Gamma(0.5625, -x^2)]+C-----(1)$

When , x=1, gives , y=9.

Evaluate the value of C and substitute in the equation 1.

6 0

3 years ago

Need help pls and thank you

Mkey [24]

Answer:

y=25x

Step-by-step explanation:

It is a linear equation with a slope of 25

5 0

2 years ago