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3 years ago

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Cecilia is making shorts with 8 belt loop. To make the loops, she needs 8 strips of fabric. Each strip measures 1 1/4 inches wid

e and 7 inches long. What is the area of each strip? Plz help and show me work

1 answer:

Korolek [52]3 years ago

6 0

Well 1/4 times 7 is 1 3/4 and 7 times 1 is 7 so add 1 3/4 and 7 to get 8 3/4 in.^2
hope i helped

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Maria had planned to walk 10 1/8 miles on wednesday, if she walked 7 2/8 miles in the morning how far would she need to walk in

Ugo [173]

Answer:

Distance Maria had to walk in the afternoon is $2\frac{7}{8}$

Step-by-step explanation:

Lets take that Maria walked $x$ miles in the afternoon.

Distance walked in the morning + distance walked in the afternoon = $10\frac{1}{8}$

⇒ $7\frac{2}{8} +x=10\frac{1}{8}$

$\frac{58}{8} +\frac{8x}{8} =\frac{81}{8}$

$\frac{58+8x}{8} =\frac{81}{8}$

$58+8x=81$

$8x=81-58\\$

$8x=23$

$x=\frac{23}{8}$

$x=2\frac{7}{8}$

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3 years ago

For the arthritic sequence 5,9,13,17,21,... how many terms would you need to get to 129

Vlad1618 [11]

9514 1404 393

Answer:

32

Step-by-step explanation:

The first term is 5 and the common difference is 9-5=4, so the n-th term is ...

an = a1 +d(n -1)

an = 5 +4(n -1) = 4n +1

Then the term number for the term 129 is ...

129 = 4n +1 . . . . . put the given value in the formula

128 = 4n . . . . . . . subtract 1

32 = n . . . . . . . . . .divide by 4

The 32nd term is 129.

8 0

3 years ago

10 people in a room each person shakes hands with every other person in the room exactly once how many handshakes will there be

Alex73 [517]

There will be 100 hand shakes because you would do 10x10 which is 100

4 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP!!!!

mezya [45]

64, because its square root is 8.
49, because its square root is 7.
144, because its square root is 12.
4, because its square root is 2.

3 0

3 years ago

Read 2 more answers

Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E

Ipatiy [6.2K]

Answer:

(A) $y=ke^{2t}$ with $k\in\mathbb{R}$ .

(B) $y=ke^{2t}/2-1/2$ with $k\in\mathbb{R}$

(C) $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$

(D) $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ ,

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then $0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}$ . With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form $e^{2t}$ with $t$ real.

(B) Proceeding and the previous item, we obtain $1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2$ . Which is not a vector space with the usual operations (this is because $-1/2$ ), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set $y=e^{mt} \Rightarrow y''=m^2e^{mt}$ and we obtain the characteristic equation $0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$ , and as in the first items the set of solutions form a vector space.

(D) Using C, let be $y=me^{3t}$ we obtain that it must satisfies $3^2m-4m=1\Rightarrow m=1/5$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ , and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).

4 0

3 years ago