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marin [14]
2 years ago
9

Solve for y -5y+8=-3y+10 what is the answer

Mathematics
2 answers:
masha68 [24]2 years ago
8 0
-5y + 8 = -3y + 10
We need to get y on one side.
Add 3y to both sides.
-2y + 8 = 10
We need to get 2y by itself.
Subtract 8 from both sides.
-2y = 2
Divide both sides by -2 to solve for y.
y = -1

I hope this helps!
AysviL [449]2 years ago
4 0
Move like terms to each side and simplify.
-5y+8=-3y+10 \\ 
-3y+5y=8-10 \\ 2y=-2 \\ y=-1
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Ludmilka [50]
Answer:
v = 18

Explain:
Equation; 4v = 72

Begin by dividing both sides
by 4 and isolating the v
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hopefully this helps you with your problem!
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5 0
3 years ago
Convert the complex number z = 4 - 10i from rectangular form to polar form.​
mrs_skeptik [129]

ANSWER

r = 2 \sqrt{29} ( \cos(292 \degree)  + \sin(292 \degree))

EXPLANATION

The polar form of a complex number ,

z = x + yi

is given by:

z = r( \cos( \theta)  + i \sin( \theta) )

where

r =  \sqrt{ {x}^{2} +  {y}^{2}  }

The given complex number is:

z = 4 - 10i

r =  \sqrt{ {4}^{2}  +  {( - 10)}^{2} }

r =  \sqrt{16  + 100}

r =  \sqrt{116}  = 2 \sqrt{29}

And

\theta=  \tan^{ - 1}(\frac{ y}{x} )

\theta=  \tan^{ - 1}(\frac{  - 10}{4} )

\theta=  292 \degree

Hence the polar form is :

r = 2 \sqrt{29} ( \cos(292 \degree)  + \sin(292 \degree))

6 0
3 years ago
Which value of www makes 14=11+\dfrac{w}{8}\cdot614=11+ 8 w ​ ⋅614, equals, 11, plus, start fraction, w, divided by, 8, end frac
Liono4ka [1.6K]

Answer:

Option B) w = 4

Step-by-step explanation:

We have to find the value of w to make the given expression true.

The given expression is:

14=11+\dfrac{w}{8}\cdot6\\\\14-11 = \dfrac{w}{8}\cdot6\\\\3 = \dfrac{w}{8}\cdot6\\\\\Rightarrow w = \dfrac{3 \cdot 8}{6} = 4

Option B) w = 4 is the correct answer.

8 0
2 years ago
Read 2 more answers
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AlexFokin [52]

Answer:

I think the answer is A

Step-by-step explanation:

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4 0
3 years ago
If one quart is the same amount as 2 pints, how many are in 8 quearts?
Verizon [17]

Answer

16 pints

Step-by-step explanation:

its basically 8x2

6 0
2 years ago
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