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3 years ago

16.827 to the nearest whole.

Mathematics

1 answer:

Hatshy [7]3 years ago

7 0

Answer:

Step-by-step explanation:

We want to round 16.827 to the nearest whole number.

When we round a number to the nearest place value, we always look at the digit in the place value to its right.

Here, we're technically rounding to the nearest ones place, so look at the tenths place. It's an 8. Remember that we round up if the digit is 5-9 and round down if the digit is 0-4.

Since it's 8, we round up. Thus, 16.827 rounds up to 17.

<em>~ an aesthetics lover</em>

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Alecsey [184]

Answer:

Step-by-step explanation:

we could use inverse operation and multiply 7 by 6

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3 years ago

Ariel wants to write 8 entries for her blog. After 2 hours, she has 5 entries left. After 4 hours, she has 2 entries left. Which

mash [69]

Let x be a time in hours and y be an amount of remaining entries. You have that:

At start Ariel wants to write 8 entries for her blog. This means that x=0 and y=8 (0 entries were written). The point with coordinates (0,8) belongs to needed graph.
After 2 hours, she has 5 entries left. This gives you that at x=2, y=5 and graph passes through the point (2,5).
After 4 hours, she has 2 entries left. This gives you that at x=4, y=2 and graph passes through the point (4,2).

Find the slope of the line that is passing through the points $(x_1,y_1)$ and $(x_2,y_2)$ by formula

$\dfrac{y_2-y_1}{x_2-x_1}.$

Thus, the slope is

$\dfrac{5-2}{2-4}=-1.5.$

Answer: graph of line going through (0, 8) with a slope of -1.5

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4 years ago

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Alla [95]

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2 years ago

(15 pts) 4. Find the solution of the following initial value problem: y"-10y'+25y = 0 with y(0) = 3 and y'(0) = 13

jolli1 [7]

Answer:

$y(x)=3e^{5x}-2xe^{5x}$

Step-by-step explanation:

The given differential equation is $y''-10y'+25y=0$

The characteristics equation is given by

$r^2-10r+25=0$

Finding the values of r

$r^2-5r-5r+25=0\\\\r(r-5)-5(r-5)=0\\\\(r-5)(r-5)=0\\\\r_{1,2}=5$

We got a repeated roots. Hence, the solution of the differential equation is given by

$y(x)=c_1e^{5x}+c_2xe^{5x}...(i)$

On differentiating, we get

$y'(x)=5c_1e^{5x}+5c_2xe^{5x}+c_2e^{5x}...(ii)$

Apply the initial condition y (0)= 3 in equation (i)

$3=c_1e^{0}+0\\\\c_1=3$

Now, apply the initial condition y' (0)= 13 in equation (ii)

$13=5(3)e^{0}+0+c_2e^{0}\\\\13=15+c_2\\\\c_2=-2$

Therefore, the solution of the differential equation is

$y(x)=3e^{5x}-2xe^{5x}$

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3 years ago

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tiny-mole [99]

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3 years ago