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3 years ago

What is the distance from point D to point C

Mathematics

1 answer:

In-s [12.5K]3 years ago

4 0

Answer:

-4to 2

Step-by-step explanation:

it lies in the x axis

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Can someone help me with number 10, 11, 12, 15, and 16 I don’t understand them thanks

yanalaym [24]

10)

$10^2-48:6+25*3=100-8+75=167$

11)

$8(\frac{16}{4} )+2^2-11*3=32+4-33=3$

12)

$(\frac{6}{3} +4)^2:4*7=(2+4)^2:4*7=6^2:4*7=36:4*7=9*7=63$

13) (your answer is incorrect)

$5(9-4)^2-3^2=5*5^2-3^2=5*25-9=125-9=116$

14) (your answer is incorrect)

$5^2-2^2*4^2-12=25-4*16-12=25-64-12=-51$

15)

$(\frac{50}{5^2} )^2:4=(\frac{50}{25} )^2:4=2^2:4=4:4=1$

16)

$\frac{24+32+30+28}{2}$ -expression to represent this situation

$\frac{24+32+30+28}{2}=\frac{114}{2}=57$

Answer: In 1 group 57 students

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P.S. Hello from Russia :^)

4 0

3 years ago

Andre drives a bike for 4 hours covering 160 km. Find the distance covered by him in 7 hours.

swat32

Answer:

I think is 280km :))))))

4 0

2 years ago

Read 2 more answers

What is Best Buy out of 3 lbs for 1.89 or 5lbs for 2.49

vodomira [7]

1.89 / 3 = 0.63 per pound
2.49/ 5 = 0.498 per pound

answer
<span>5lbs for 2.49 is the best buy</span>

8 0

3 years ago

Read 2 more answers

If 3m - n = 11, then m =

irina [24]

Answer:

n= -2 and m=3

Step-by-step explanation:

3 x 3= 9

11- 9 = 2

5 0

3 years ago

Please I need help with differential equation. Thank you

Inga [223]

1. I suppose the ODE is supposed to be

$\mathrm dt\dfrac{y+y^{1/2}}{1-t}=\mathrm dy(t+1)$

Solving for $\dfrac{\mathrm dy}{\mathrm dt}$ gives

$\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{y+y^{1/2}}{1-t^2}$

which is undefined when $t=\pm1$ . The interval of validity depends on what your initial value is. In this case, it's $t=-\dfrac12$ , so the largest interval on which a solution can exist is $-1\le t\le1$ .

2. Separating the variables gives

$\dfrac{\mathrm dy}{y+y^{1/2}}=\dfrac{\mathrm dt}{1-t^2}$

Integrate both sides. On the left, we have

$\displaystyle\int\frac{\mathrm dy}{y^{1/2}(y^{1/2}+1)}=2\int\frac{\mathrm dz}{z+1}$

where we substituted $z=y^{1/2}$ - or $z^2=y$ - and $2z\,\mathrm dz=\mathrm dy$ - or $\mathrm dz=\dfrac{\mathrm dy}{2y^{1/2}}$ .

$\displaystyle\int\frac{\mathrm dy}{y^{1/2}(y^{1/2}+1)}=2\ln|z+1|=2\ln(y^{1/2}+1)$

On the right, we have

$\dfrac1{1-t^2}=\dfrac12\left(\dfrac1{1-t}+\dfrac1{1+t}\right)$

$\displaystyle\int\frac{\mathrm dt}{1-t^2}=\dfrac12(\ln|1-t|+\ln|1+t|)+C=\ln(1-t^2)^{1/2}+C$

$2\ln(y^{1/2}+1)=\ln(1-t^2)^{1/2}+C$

$\ln(y^{1/2}+1)=\dfrac12\ln(1-t^2)^{1/2}+C$

$y^{1/2}+1=e^{\ln(1-t^2)^{1/4}+C}$

$y^{1/2}=C(1-t^2)^{1/4}-1$

I'll leave the solution in this form for now to make solving for $C$ easier. Given that $y\left(-\dfrac12\right)=1$ , we get

$1^{1/2}=C\left(1-\left(-\dfrac12\right)^2\right))^{1/4}-1$

$2=C\left(\dfrac54\right)^{1/4}$

$C=2\left(\dfrac45\right)^{1/4}$

and so our solution is

$\boxed{y(t)=\left(2\left(\dfrac45-\dfrac45t^2\right)^{1/4}-1\right)^2}$

3 0

3 years ago