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sweet [91]
3 years ago
6

Which value is not a solution of the inequality -x + 5 < 7? Select all

Mathematics
1 answer:
ehidna [41]3 years ago
3 0

Answer:

-3

-12

-5

Step-by-step explanation:

A negative number with another negative symbol put onto it is positive because both negative symbols cancel each other out

So -(-3) = 3

-(-12) = 12

-(-5) = 5

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Practice: triangle sum theorem on a piece of paper solve for the variable , then find the missing angles
MariettaO [177]
1) x=-3. The missing angle is 71 degrees
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3 years ago
2/5 k - 3/5 + 1/10 k
liraira [26]

Answer:

Step-by-step explanation:

2/ 5 k− 3 /5 k+ 1 /10 k = 2 /5 k+ −3 /5 k+ 1 /10 k

Combine Like Terms:

= 2 /5 k+ −3 /5 k+ 1 /10 k =( 2 /5 k+ −3 /5 k+ 1 /10 k)

= −1 /10 k

Answer:

= −1 /10 k

8 0
3 years ago
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8 yd 15 yd 4 yd r ay Find the volume of the prism. Round your answer to the nearest tenth. 18.5 ft 8.6 ft h = 8.4 ft​
seraphim [82]
  • L=8yd
  • B=15yd
  • H=4yd

Volume:-.

\\ \tt\longmapsto V=LBH

\\ \tt\longmapsto V=8(15)(4)

\\ \tt\longmapsto V=480yd^3

6 0
2 years ago
Simplify the following expressions and remember to pay attention to the signs plz help me
ioda

2 negatives can be converted to a plus sign. -(-x) can be +x or just x.

-54--23

-54+23

<u>23-54=-31</u>

78--4

<u>78+4=82</u>

-48/-6

Now let's pause fer a sec. When a negative is being divided by another negative, the quotient(answer to division problem) becomes a positive.

<u>-48/-6=8</u>

<u></u>

<u>50*-4=-200</u>

---

hope it helps

7 0
3 years ago
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Let v be an eigenvector of a matrix A with a corresponding eigenvalue ?=2. Find one solution x of the system Ax=v.
Murljashka [212]

Answer with explanation:

For, a Matrix A , having eigenvector 'v' has eigenvalue =2

 The order of matrix is not given.

It has one eigenvalue it means it is of order , 1×1.

→A=[a]

Determinant [a-k I]=0, where k is eigenvalue of the given matrix.

It is given that,

k=2

For, k=2, the matrix [a-2 I] will become singular,that is

→ Determinant |a-2 I|=0

→I=[1]

→a=2

Let , v be the corresponding eigenvector of the given eigenvalue.

→[a-I] v=0

→[2-1] v=[0]

→[v]=[0]

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Now, corresponding eigenvector(v), when eigenvalue is 2 =0

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→Ax=v

→[2] x=0

→[2 x] =[0]

→x=0, is one solution of the system.

5 0
4 years ago
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