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defon
3 years ago
8

Please help, I dont know where to start​

Mathematics
1 answer:
PtichkaEL [24]3 years ago
4 0

Answer:

16 ft

Step-by-step explanation:

The circumference of a circle is given by

C = pi d

where d is the diameter

50.24 = 3.14 *d

Divide each side by 3.14

50.24/ 3.14 = 3.14d/3.14

16 = d

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Matrices A and B are square matrices of the same size. Prove Tr(c(A + B)) = C (Tr(A) + Tr(B)).
alexira [117]

Answer with Step-by-step explanation:

We are given that two matrices A and B are square matrices of the same size.

We have to prove that

Tr(C(A+B)=C(Tr(A)+Tr(B))

Where C is constant

We know that tr A=Sum of diagonal elements of A

Therefore,

Tr(A)=Sum of diagonal elements of A

Tr(B)=Sum of diagonal elements of B

C(Tr(A))=C\cdot Sum of diagonal elements of A

C(Tr(B))=C\cdot Sum of diagonal elements of B

C(A+B)=C\cdot (A+B)

Tr(C(A+B)=Sum of diagonal elements of (C(A+B))

Suppose ,A=\left[\begin{array}{ccc}1&0\\1&1\end{array}\right]

B=\left[\begin{array}{ccc}1&1\\1&1\end{array}\right]

Tr(A)=1+1=2

Tr(B)=1+1=2

C(Tr(A)+Tr(B))=C(2+2)=4C

A+B=\left[\begin{array}{ccc}1&0\\1&1\end{array}\right]+\left[\begin{array}{ccc}1&1\\1&1\end{array}\right]

A+B=\left[\begin{array}{ccc}2&1\\2&2\end{array}\right]

C(A+B)=\left[\begin{array}{ccc}2C&C\\2C&2C\end{array}\right]

Tr(C(A+B))=2C+2C=4C

Hence, Tr(C(A+B)=C(Tr(A)+Tr(B))

Hence, proved.

5 0
3 years ago
Please help me out with this. God bless
Burka [1]
The answer is 22.6 for the top one and 6x+14y+13 for the second one... the coefficient for y=14
4 0
3 years ago
In a survey, 15 high school students said they could drive and 15 said they could not. Out of 60 college students surveyed, 30 s
Citrus2011 [14]

Answer:

No.

Step-by-step explanation:

#1, the sample size is not indicative of a proper high school population, unless of all there is a small amount of students in the high school (or even only 30 students in it). On a individual basis, it can vary widely, and so data analysis such as this is not correct. The data size is skewed as well for the college, especially as college campuses typically house more students then usual. To only ask 60 college students also skew the data gotten. Another sample error may be locational error, in which the location of the schools may warrant the need for a driver license, or if it is not needed, as well as the distance of the student's house and/or where needed facilities are to accommodate a student's needs and desires.

If we were to set all these problems aside and focus more on the question, 15 high school students said they could drive, and 15 said they could not, which makes the sample size of 30 high school students in all. 15 say they can drive, therefore:

15/30 = 1/2, or there is a 50% chance of obtaining a student that can drive.

On the other hand, out of 60 college students surveyed, 30 said they can drive, therefore:

30/60 = 1/2, or there is a 50% chance of obtaining a student that can drive.

∴ Micah's prediction is incorrect in obtaining a college student that can drive more easily then obtaining one that can drive in a high school.

However, remember that too small of a data size can lead to a heavy skew, and it will not represent reality as well.

~

8 0
3 years ago
Select the correct answer.
TEA [102]
66 degrees

Add 30, 57, and 27, then subtract from 180
7 0
3 years ago
Read 2 more answers
Calculate the area of the shaded region.
Nostrana [21]

Let A be the area of the sector in the larger circle (radius 12 cm) whose central angle is subtended by the labeled arc with measure 45 deg, and let a be the area of the sector in the smaller circle (radius 8 cm) with the same central angle. The area you want is A-a.

We have

\dfrac A{\pi(12\,\mathrm{cm})^2}=\dfrac{45^\circ}{360^\circ}\implies A=18\pi\,\mathrm{cm}^2

\dfrac a{\pi(8\,\mathrm{cm})^2}=\dfrac{45^\circ}{360^\circ}\implies a=8\pi\,\mathrm{cm}^2

So the area of the shaded region is 18\pi-8\pi=10\pi.

7 0
3 years ago
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