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snow_lady [41]
3 years ago
11

(4x + 8) Solve for the value of x in the picture above.

Mathematics
1 answer:
FinnZ [79.3K]3 years ago
5 0

Answer: C

Step-by-step explanation:

The triangle in the picture is an equilateral triangle. We know this is true because all sides are equal, as denoted by the tick marks. In equilateral triangles, each angle is 60° because when you add all the angles, you get 180°.

4x+8=60        [subtract both sides by 8]

4x=52            [divide both sides by 4]

x=13

Now, we know that C is the correct answer.

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Another math problem help please?
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Circles M and K are congruent, arc QR is congruent to arc LN and arc OP is congruent to arc VW. Find x.
SCORPION-xisa [38]

Given, arc QR is congruent to arc LN and arc OP is congruent to arc VW.

And the expressions for each arc in the diagram also given as:

Arc QR = 2x - y, arc LN = 11 , arc OP= 10 and arc VW=5x+y.

Hence, we will get the system of equations as following:

Arc QR = Arc LN

2x - y = 11 ...(1)

Arc OP = Arc VW

5x + y = 10 ...(2)

We need to find the value of x. So, we can add the equations to eliminate y so that we can solve the equations for x. Therefore,

2x+5x = 11 + 10

7x = 21

\frac{7x}{7} =\frac{21}{7} Divide each sides by 7.

So, x= 3

6 0
3 years ago
mr.rivera bought some erasers to give his students. on monday , mr.rivera gave out 22 erasers. on Tuesday, he gave out 39 earser
Mumz [18]

Answer:

178 ERASERS

Step-by-step explanation:

YOU NEED TO START BACKWARDS. HE BOUGHT 76 ERASERS AND HE HAS 193, SO WHATEVER HE HAD AT THE END OF TUESDAY+76 IS HOW MUCH HE HAS ON WEDNESDAY. 193-76=117. ON TUESDAY, HE STARTED WITH ERASERS AT END OF TUESDAY+39 ERASERS. 117+39=156. ON MONDAY, HE STARTED WITH END OF MONDAY+22 ERASERS. 156+22=178

6 0
3 years ago
R = sec(θ) − 2cos(θ), where -π/2 < θ < π/2
Alex

Answer:

  y = (x/(1-x))√(1-x²)

Step-by-step explanation:

The equation can be translated to rectangular coordinates by using the relationships between polar and rectangular coordinates:

  x = r·cos(θ)

  y = r·sin(θ)

  x² +y² = r²

__

  r = sec(θ) -2cos(θ)

  r·cos(θ) = 1 -2cos(θ)² . . . . . . . . multiply by cos(θ)

  r²·r·cos(θ) = r² -2r²·cos(θ)² . . . multiply by r²

  (x² +y²)x = x² +y² -2x² . . . . . . . substitute rectangular relations

  x²(x +1) = y²(1 -x) . . . . . . . . . . . subtract xy²-x², factor

  y² = x²(1 +x)/(1 -x) = x²(1 -x²)/(1 -x)² . . . . multiply by (1-x)/(1-x)

  \boxed{y=\dfrac{x\sqrt{1-x^2}}{1-x}}

__

The attached graph shows the equivalence of the polar and rectangular forms.

4 0
2 years ago
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