All categories

3 years ago

Help please Will mark ya as brainliest for most accurate answer plzz

Mathematics

2 answers:

just olya [345]3 years ago

7 0

Answer:

57.06 unit^2

Step-by-step explanation:

find the area of the rectangle = 12*12 = 144 unit^2

the area of the circle = pi*r^2 = 3.14*8^2 = 201.06 unit^2 (rounded to the nearest hundredth)

then the "shaded area" is the difference

201.06 - 144 = 57.06 unit^2

leva [86]3 years ago

5 0

Answer w/o rounding: 176.96
Answer with rounding:177

You might be interested in

HELP ASAP.<br><br> please help the best answer gets brainliest

Aleonysh [2.5K]

Answer:

She should have cross multiplied.

Step-by-step explanation:

-2.125, or 17/8, is the correct answer

4 0

3 years ago

Read 2 more answers

At Jefferson middle school 82 students were asked which sport they plan to participate in for the upcoming year. 20 students pla

worty [1.4K]

Answer:

2 kids plan to do cross country

Step-by-step explanation:

8 0

3 years ago

Event v occurs 28% of the time on Tuesdays.

vampirchik [111]

Answer:

68%

Step-by-step explanation:

Probability of occurrence of Event v = P(v) = 28% = 0.28

Probability of occurrence of both Events v and Event w together = P(v and w) = 19% = 0.19

We have to find what is the probability that event w occurs with event v given that event v occurs on a Tuesday. This is a conditional probability. In other words we have to find what is the probability of event w given that event v occurs of Tuesday. i.e we have to find P(w|v)

The formula to calculate this conditional probability is:

$P(w|v) = \frac{P(v \cap w)}{P(v)}$

Using the given values, we get:

$P(w|v) = \frac{0.19}{0.28}\\\\ P(w|v) = 0.68\\\\ P(w|v) = 68\%$

Therefore, the probability that even w will occur with event v given that event v occurs on Tuesday is 68%

7 0

3 years ago

Find the slope of a line given two or more points (2,4)(5,3)

masha68 [24]

Answer:

-1/3 (can u plz give me the brainliest answer???)

Step-by-step explanation:

change in y over the change in x. (or rise over run).

-1/3

Have a nice day!!!!!!!!!!!

7 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 $k\in\mathbb{R}$

(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