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attashe74 [19]
3 years ago
9

Any suggestions or answers

Mathematics
2 answers:
S_A_V [24]3 years ago
6 0

Answer:

c)x=7

Step-by-step explanation:

a) MNK=RTP

b)NM

c)x=?

20=3x-1

3x=21

x=7

Bumek [7]3 years ago
6 0
MNK is congruent to RTP
TR is congruent to NM
solve for x:
3x-1=20
add 1 to both sides
3x=21
divide both sides by 3
x=7
x=
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Step 4 is where she made the error.

Step-by-step explanation:

This step is actually known as the commutative property. That property states that we can change the order of numbers as long as we keep the signs and it maintains it's validity.

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Simplify. Show your work. <br><br> ( 6x-2 ) ^2 ( 0.5x )^4
mrs_skeptik [129]

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About how many u.s dollars would you get if you exchange 15 euros
Free_Kalibri [48]

Answer:

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Step-by-step explanation:

8 0
3 years ago
How do I solve for x?
notka56 [123]
Solving:
\frac{3+4x}{2} + \frac{5-x}{3} = \frac{29}{6}
Make the Least Common Multiple (2,3,6)
2,3,6\:|2
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1,1,1\:|\underline{2*3=6}


<span>Replace denominators and resolve:
</span>\frac{3+4x}{2} + \frac{5-x}{3} = \frac{29}{6}
\frac{3(3+4x)}{6} + \frac{2(5-x)}{6} = \frac{29}{6}
Cancel the dominators
\frac{3(3+4x)}{\diagup\!\!\!\!6} + \frac{2(5-x)}{\diagup\!\!\!\!6} = \frac{29}{\diagup\!\!\!\!6}
3(3+4x) + 2(5-x) = 29
9 + 12x + 10 - 2x = 29
12x - 2x = 29 - 9 - 10
10x = 20 - 10
10x = 10
x =  \frac{10}{10}
\boxed{\boxed{x = 1}}\end{array}}\qquad\quad\checkmark


7 0
3 years ago
PLZ HELP !! Which of the points (s, t) is not one of the vertices of the shaded region of the
stira [4]

Answer:

A(0,10)

Step-by-step explanation:

Given the 4 inequalities:

s ≥ 12 - 0.5t (1)

s ≥ 10 -t (2)

s ≤ 20-t (3)

s ≥ 0

t ≥ 0

Let analyse all 4 possible answer:

  • A(0, 10)

Let substitute this point into (1) we have: 10 ≥ 12 -0.5*0 Wrong

We do not choose A

  • B (0, 20)

Let substitute this point into (1) we have: 20 ≥ 12 -0.5*0  True

Let substitute this point into (2) we have: 20 ≥ 10 - 0 True

Let substitute this point into (3) we have: 20 ≤ 20 - 0 True

We choose B as the vertex

  • C. (4, 16)

Let substitute this point into (1) we have: 16 ≥ 12 -0.5*4  True

Let substitute this point into (2) we have: 16 ≥ 10 - 4 True

Let substitute this point into (3) we have: 16 ≤ 20 - 4 True

We choose C as the vertex

  • D. (16,4)

Let substitute this point into (1) we have: 4 ≥ 12 -0.5*16 True

Let substitute this point into (2) we have: 4 ≥ 10 - 16 True

Let substitute this point into (3) we have: 4 ≤ 20 - 16 True

We choose B as the vertex

Hence, the point A(0,10)  is not one of the vertices of the shaded region of the  set of inequalities.

3 0
3 years ago
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