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schepotkina [342]
3 years ago
7

Find the value of x.

Mathematics
1 answer:
shepuryov [24]3 years ago
8 0

The two given equations are vertical angles, which mean they are the same.

Set the two equations to equal and solve for x:


13x-23 = 5x+17

Add 23 to each side:

13x = 5x+40

Subtract 5x from each side:

8x = 40

Divide both sides by 8:

x = 40/8

x = 5



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Answer:

358.9 I think

Step-by-step explanation:

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2 years ago
Which property is used in this statement? if k/2=5, then k=10
Otrada [13]
Property of equality
k=k
10/2=5
5=5
3 0
3 years ago
Ryan has some donuts he takes 5 away <br><br> d - 5 = ?
Serga [27]

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2

Step-by-step explanation:

6 0
3 years ago
Over the interval [0, 2π), what are the solutions to cos(2x) = cos(x)? Check all that apply.
valentinak56 [21]

Answer:

Step-by-step explanation:

given the expression;

cos(2x) = cos(x)

According to trig identity;

cos(2x) = cos(x+x)

cos(2x) = cos x cos x - sinx sinx

cos(2x) = cos²(x)-sin²(x)

cos(2x) = cos²(x)-(1-cos²x)

cos(2x) = cos²(x)+cos²x-1

cos(2x) = 2cos²(x)-1

2cos²(x)-1 = cos(x)

let P = cosx

2P²-1 = P

2P²-P-1 = 0

Factorize;

2P²-2P+P-1 = 0

2P(P-1)+1(P-1) = 0

2P - 1 = 0 and P-1 = 0

P = 1/2 and 1

cosx = 1/2 and cos x = 1

x = arccos 1/2

x = π/3

Also;

x = arccos1

x = 0

Hence the value of x are 0 and  π/3

Also the angle = π+ π/3 = 4π/3

The angles are 0, π/3 and 4π/3

4 0
3 years ago
Read 2 more answers
I need help with c please.
STALIN [3.7K]

Answer:

||w|| = 6

Step-by-step explanation:

||w|| = modulus or magnitude of vector w

Since formula to get the modulus of any vector = \sqrt{\text{(x-component)}^2+\text{(y-component)}^2}

Vector w = <-6, 0>

x-component of vector w = (-6)

y-component of vector w = 0

Therefore, ||w|| = \sqrt{(-6)^2)+(0)^2}

                         = 6

3 0
3 years ago
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