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Aleonysh [2.5K]
3 years ago
8

3x - 9 = 12 Solve for X

Mathematics
2 answers:
Flauer [41]3 years ago
8 0
Okay, simple.

x=7 

hope this helps

kozerog [31]3 years ago
5 0
<span>3x - 9 = 12 
3x = 12 + 9
3x = 21
x =21/3
x =7
hope that helps</span>
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4 years ago
For a field trip, 6 students rode in cars and the rest
Marrrta [24]

Answer:

(326-6) ÷ 8 = s

40 = s

Step-by-step explanation:

326-6=320 This was done so we get to know the total amount of students in busses, since six rode in cars.

Then we divide the amount of students in busses by the amount of busses to get 40. 40 students in each bus.

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3 years ago
A sewing machine is at a 20% off sale. The original price was $340. How much does the sewing machine cost at a 20% off sale?
Anvisha [2.4K]

Answer:

272$

Step-by-step explanation:

Since it is at a sale of 20% off that means that we will have to subtract 100% from 20% which is 80%

Now to find the exact value, we will have to multiply 80%, or 0.80 with 340

0.80*340= 272$

Therefore our answer is 272$

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4 0
3 years ago
Read 2 more answers
WILL MARK BRANIEST: If the magnitude of the block's acceleration is 1.5 m/s2, what is the mass of the block? Show your work. (Th
puteri [66]

Answer:Answer:

F = 0.84 [N]

Explanation:

In order to solve this problem we must take into account Newton's second law, which tells us that the sum of all forces applied on a body must be equal to the product of the mass of the body by its acceleration.

F = m*a

where:

F = force [N]

a = acceleration = 1.4[m/s^2]

m = mass = 0.6 [kg]

F = 0.6*1.4

F = 0.84 [N]

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3 years ago
How to find left and right inverse of a 3x2 matrix.
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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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