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sergey [27]
3 years ago
15

Round 141080 to the nearest ten

Mathematics
2 answers:
Svetllana [295]3 years ago
8 0
141080 is my answer because when you look at the second number from the right, it's 8 and the number next to it is 0 so numbers below 5 round down so that is my answer
Licemer1 [7]3 years ago
3 0
141080 its already rounded to the nearest ten
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Four times the sum of g+6 is equal to 36 what is the value of g?
solniwko [45]

Answer:

g=3

Step-by-step explanation:

4(g+6)=36

4g+24=36

   -24   -24

4g=12

/4   /4

g=3

Check:

4(3+6)=36

4*9=36

36=36

7 0
3 years ago
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You roll two standard number cubes. What is the probability that the sum is odd,
Free_Kalibri [48]
The correct answer would be slightly over 33%.<span />
7 0
3 years ago
What are the vertex and x-intercepts of the graph of the function below?
geniusboy [140]

y= x^2 - 6x-7

Vertex = -b /2a

b= -6

a= 1

-(-6)/2(1)= 3

x= 3

Use the substitution method to find y

y= (3)^2-6(3)-7

(3)(3)-18-7

9-25

y= -16

Vertex is (3, 16)

Find x- intercept (y= 0)

y= x^2 - 6x-7

(x-7)(x+1)

x-7= 0

Move -7 to the other side. Sign changes from -7 to 7

x-7+7= 0+7

x= 7

x+1= 0

Move +1 to the other side. Sign changes from +1 to -1

x+1-1= 0-1

x= -1

Answer :B. Vertex: (3, -16); Intercepts: x=-1,7

5 0
3 years ago
Which expression is equivalent to 3(s+7)?<br> 21s<br> 10s<br> 3s+21<br> 21s+3
VMariaS [17]
You have to distribute so the correct answer would be 3s+21
6 0
3 years ago
Read 2 more answers
How to find left and right inverse of a 3x2 matrix.
Mnenie [13.5K]

Let A be a 3×2 matrix, L its left inverse, and R its right inverse. L and R are then matrices such that LA = I₂ (the 2×2 identity matrix) and AR = I₃ (the 3×3 identity matrix). Clearly L must be 2×3 and R must be 3×2 in order for the matrix products to be defined.

To find L and R, we start by introducing a square matrix on the the left sides of either equation above. In particular, we uniformly multiply both sides by the transpose of A, then solve for the inverse.

For the left inverse, we have

LA=I

(LA)A^\top = IA^\top

L\left(AA^\top\right) = A^\top

\left(L\left(AA^\top\right)\right)\left(AA^\top\right)^{-1} = A^\top \left(AA^\top\right)^{-1}

L\left(\left(AA^\top\right)\left(AA^\top\right)^{-1}\right) = A^\top \left(AA^\top\right)^{-1}

LI = A^\top \left(AA^\top\right)^{-1}

L = A^\top \left(AA^\top\right)^{-1}

We do the same thing for the right inverse, but take care with how we multiply both sides of AR = I₃.

AR=I

A^\top(AR)=A^\top I

\left(A^\top A\right)R = A^\top

\left(A^\top A\right)^{-1} \left(\left(A^\top A\right)R\right) = \left(A^\top A\right)^{-1} A^\top

\left(\left(A^\top A\right)^{-1} \left(A^\top A\right)\right) R = \left(A^\top A\right)^{-1} A^\top

IR = \left(A^\top A\right)^{-1} A^\top

R = \left(A^\top A\right)^{-1} A^\top

4 0
3 years ago
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