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3 years ago

13

What is 80% of 80 round to the nearest hundreds

2 answers:

emmainna [20.7K]3 years ago

8 0

Answer:64

Step-by-step explanation:

8090 [49]3 years ago

6 0

512,000 80 times 100 times 80 times .8

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Three times a number increased by 20, is 14 less than the product of five and the 1095

Zinaida [17]

$3x + 20 = (5 \times 1095) - 14 \\ 3x + 20 = 5475 - 14 \\ 3x = 5461 - 20 \\ 3x = 5441 \\ x = \frac{5441}{3}$

8 0

2 years ago

G(n)=3n-1 <br> H(n)=n^2-3<br> Find (g*h)(-4)

IrinaVladis [17]

Answer:

Step-by-step explanation:

G(-4)=3(-4)-1=-13

H(-4)=(-4)^2-3=16-3=13

(G*H)(-4)=G(-4)*H(-4)=-13*13=-169

7 0

3 years ago

This morning, Megan and her dad had smoothies for breakfast. Megan made her smoothie with 1 cup of mango and 3 cups of strawberr

murzikaleks [220]

Answer:

Yes.

Step-by-step explanation:

4 0

3 years ago

Use the variation of parameters method to solve the DR y" + y' - 2y = 1

postnew [5]

Answer:

$y(t)\ =\ C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

Step-by-step explanation:

As given in question, we have to find the solution of differential equation

$y"+y'-2y=1$

by using the variation in parameter method.

From the above equation, the characteristics equation can be given by

$D^2+D-2\ =\ 0$

$=>D=\ \dfrac{-1+\sqrt{1^2+4\times 2\times 1}}{2\times 1}\ or\ \dfrac{-1-\sqrt{1^2+4\times 2\times 1}}{2\times 1}$

$=>\ D=\ -2\ or\ 1$

Since, the roots of characteristics equation are real and distinct, so the complementary function of the differential equation can be by

$y_c(t)\ =\ C_1e^{-2t}+C_2e^t$

Let's assume that

$y_1(t)=e^{-2t}$ $y_2(t)=e^t$

$=>\ y'_1(t)=-2e^{-2t}$ $y'_2(t)=e^t$

and g(t)=1

Now, the Wronskian can be given by

$W=y_1(t).y'_2(t)-y'_1(t).y_2(t)$

$=e^{-2t}.e^t-e^t(-e^{-2t})$

$=e^{-t}+2e^{-t}$

$=3e^{-t}$

Now, the particular solution can be given by

$y_p(t)\ =\ -y_1(t)\int{\dfrac{y_2(t).g(t)}{W}dt}+y_2(t)\int{\dfrac{y_1(t).g(t)}{W}dt}$

$=\ -e^{-2t}\int{\dfrac{e^t.1}{3.e^{-t}}dt}+e^{t}\int{\dfrac{e^{-2t}.1}{3.e^{-t}}dt}$

$=\ -e^{-2t}\int{\dfrac{1}{3}dt}+\dfrac{e^t}{3}\int{e^{-t}dt}$

$=\dfrac{-e^{-2t}}{3}.t-\dfrac{1}{3}$

$=-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

Now, the complete solution of the given differential equation can be given by

$y(t)\ =\ y_c(t)+y_p(t)$

$=C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

5 0

3 years ago

Its sooooo easy im just stresed and my brain doe not work

harkovskaia [24]

Answer:

The answer is D. all of the above

5 0

4 years ago

Read 2 more answers