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yan [13]
3 years ago
14

In your own words, write a verbal expression for the following algebraic expression. 18p

Mathematics
1 answer:
maxonik [38]3 years ago
6 0

Step-by-step explanation:

18p is a multiplication expression in itself.

It could be written as 18 x p. I am confused on what you mean as 'verbal' so i'm gonna take a wild guess and say word form.

eighteen times p???

Sorry if it's wrong.

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Calculate the length of x.
Naya [18.7K]

Answer:

B

Step-by-step explanation:

Use the pythagorean theorem to find the hypotenuse. You can do this by squaring 12 and 11 (separately) and then adding those together. You'll get 265. However, since this is c squared and not c, you have to find the square root for the correct answer. The square root of 265 is approximately 16.28. Hope this helps!

6 0
3 years ago
Ten percent of the reds are added to twenty percent of the blues, and the total is 24. Yet the product of the number of reds and
salantis [7]
0.10r + 0.20b = 24
3r = b + 20....b = 3r - 20

0.10r + 0.20(3r - 20) = 24
0.10r + 0.60r - 4 = 24
0.70r = 24 + 4
0.70r = 28
r = 28/0.70
r = 40 <=== there are 40 reds

0.10r + 0.20b = 24
0.10(40) + 0.20b = 24
4 + 0.20b = 24
0.20b = 24 - 4
0.20b = 20
b = 20/0.20
b = 100 <=== there are 100 blues

5 0
3 years ago
M∠ROS = 20 deg 15' 40", m∠SOT = 10 ° 12' 30" m∠ROT =
nlexa [21]
<span>ROS+SOT=30 28' 10" ROS+ROT+SOT=180 ROT=180-(ROS+SOT)= 150 32' 50"</span>
7 0
3 years ago
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
HI NEED HELP SOON!!!! PLEASE
Dvinal [7]

Answer:

8/27

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
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