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2 years ago

11

Mr.price flew 4,908 from New York then he flew 3525 miles from Hawaii to Texas how much farther was mr.price trip from New York

to Hawaii than his trip from Hawaii to Texas

1 answer:

alexandr1967 [171]2 years ago

4 0

yes yes i agree completely thats the snswet answer* 25 a b c d 46:77 77/55

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A tortoise is walking in the desert. It walks at a speed of 7 meters per minute for 8.96 meters. For how many minutes does it wa

Licemer1 [7]

Answer:

1.28 minutes

Step-by-step explanation:

Given data

Speed= 7 meters per minute

Distance= 8.96 meters

We know that the expression for the Speed, Distance, and time is given as

Speed= Distance/Time

TIme= Distance/Speed

Substitute

Time= 8.96/7

Time= 1.28 minutes

7 0

3 years ago

K= m-q/c solve for m

krok68 [10]

Add q/c to the side with k and get q/c-k=m

5 0

3 years ago

Can someone help with this trig ? pls

beks73 [17]

Answer:

45°

Step-by-step explanation:

To find you answer, we take the inverse sin(√2/2). When you plug that into your calc, you get 45°.

Alternatively, if you learned and remembered your unit circle, inverse sin(√2/2) would be π/4.

5 0

2 years ago

Read 2 more answers

An= 2an-1 +5 what is the value of a5<br><br> a1 = 1

Marianna [84]

Answer:

$a_5=91$

Step-by-step explanation:

The general term formula is:

$a_n=2a_{n-1}+5$

And given first term is "1",

The general term formula means, to get a term, multiply the previous term by 2 and sum it with "5".

Lets find a_2:

$a_2=2a_{1}+5\\a_2=2(1)+5\\a_2=7$

Now, finding a_3:

$a_3=2a_2+5\\a_3=2(7)+5\\a_3=19$

Now, a_4:

$a_4=2a_3+5\\a_4=2(19)+5\\a_4=43$

Finally, finding a_5:

$a_5=2a_4+5\\a_5=2(43)+5\\a_5=91$

This is the answer.

5 0

3 years ago

Find the work done by the vector field F~ = x 2y~i + 1 3 x 3~j + xy~k along the curve of intersection C of the paraboloid z = y

Len [333]

Let $x=\cos t$ and $y=\sin t$ . Then $C$ can be parameterized by

$\vec r(t)=\cos t\,\vec\imath+\sin t\,\vec\jmath+(\sin^2t-\cos^2t)\,\vec k$

with $0\le t\le2\pi$ , and its derivative is

$\dfrac{\mathrm d\vec r}{\mathrm dt}=-\sin t\,\vec\imath+\cos t\,\vec\jmath+4\sin t\cos t\,\vec k$

Now,

$\vec F(x,y,z)=x^2y\,\vec\imath+\dfrac{x^3}3\,\vec\jmath+xy\,\vec k$

$\implies\vec F(\vec r(t))=\cos^2t\sin t\,\vec\imath+\dfrac{\cos^3t}3\,\vec\jmath+\cos t\sin t\,\vec k$

Then the work done by $\vec F$ along $C$ is

$\displaystyle\int_C\vec F(x,y,z)\cdot\mathrm d\vec r=\int_0^{2\pi}\vec F(\vec r(t))\cdot\frac{\mathrm d\vec r}{\mathrm dt}\,\mathrm dt=\int_0^{2\pi}\left(3\cos^2t\sin^2t+\frac{\cos^4t}3\right)\,\mathrm dt=\boxed{\pi}$

8 0

3 years ago