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nekit [7.7K]
3 years ago
11

Can you graph X-intercept -3,-2,-1,1 and Y-intercept 5,2,-1,-3

Mathematics
2 answers:
arlik [135]3 years ago
7 0

--Answer--

It's in the picture, desmos will help for graphing.

Gekata [30.6K]3 years ago
7 0

Answer:(-3,5)(-2,2)(-1,-1)(1,-3)

Step-by-step explanation:

ion know it feels right

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Write a polynomial in standard form with the zeros –4, 0, 1, and 4.
Reptile [31]

Step-by-step explanation:

The answer in the picture above

4 0
3 years ago
Find all the integers,b, that make up the trinomial x2 + bx + 10 factorable.
KonstantinChe [14]

By the fundamental theorem of algebra, we can write

x^2+bx+10=(x-r_1)(x-r_2)

Expanding the right hand side, we get

x^2+bx+10=x^2-(r_1+r_2)x+r_1r_2\implies\begin{cases}r_1+r_2=-b\\r_1r_2=10\end{cases}

We want b to be an integer, which means r_1,r_2 must also be integers. This means r_1,r_2 must be factors of 10. There are several possibilities:

r_1=\pm10,r_2=\pm1\implies b=-(10+1)=-11\text{ or }b=-(-10-1)=11

r_1=\pm5,r_2=\pm2\implies b=-(5+2)=-7\text{ or }b=-(-5-2)=7

So there are 4 possible values for b: -11, -7, 7, and 11.

7 0
3 years ago
Please answer this question thank you
Tanya [424]

Answer:

74

Step-by-step explanation:

What I'm getting at, is that you need to add all the hour sold cups up, and then double it because she expects to sell twice as much. 15 + 11 +5 +6 = 37, and 37 doubled is 74

6 0
2 years ago
Read 2 more answers
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
Is negative 5 a rational number
In-s [12.5K]
Its a rational number but also a integer and could never be a whole number because its a negative and whole numbers are always positive
 
7 0
3 years ago
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