Solve for x : Y=3x- 8 Y=-4x+6

(02.07)Which relationship is always true for the angles x, y, and z of triangle ABC? A triangle is shown with a leg extending pa

zysi [14]

Check out the attached image. I drew what I think your book is showing. The figure on the left is triangle ABC without any extended segments. The figure on the right has segment AB extended shown in red. This forms the exterior angle x

The rule that connects x, y and z together is the remote interior angle theorem. It says that adding two interior angles is going to be equal to the exterior angle that is not touching either interior angle. The "remote" part means "far away" so just think of the two angles that are furthest way or not touching the exterior angle in question.

In terms of algebra, the rule is
x+y = z

3 0

3 years ago

Read 2 more answers

HELP PLZZZZZZZZ HELP PLZZZZZZZZZ

Lesechka [4]

1. 6.73 x 10^6
2. 1.33 x 10^4
3.9.77 x 10^22
4. 3.84 x 10^5
The rest i didn’t understand what where you trying to say

7 0

3 years ago

X + y = -3 & 5x - 2y = -50 ELIMINATION

Karolina [17]

I gotchu luh bro....

7 0

3 years ago

Convert the Cartesian equation (x 2 + y 2)2 = 4(x 2 - y 2) to a polar equation.

SashulF [63]

ANSWER

${r}^{2} = 4 \cos2\theta$

EXPLANATION

The Cartesian equation is

${( {x}^{2} + {y}^{2} )}^{2} = 4( {x}^{2} - {y}^{2} )$

We substitute

$x = r \cos( \theta)$

$y = r \sin( \theta)$

and

${x}^{2} + {y}^{2} = {r}^{2}$

This implies that

${( {r}^{2} )}^{2} = 4(( { r \cos\theta) }^{2} - {(r \sin\theta) }^{2} )$

Let us evaluate the exponents to get:

${r}^{4} = 4({ {r}^{2} \cos^{2}\theta } - {r}^{2} \sin^{2}\theta)$

Factor the RHS to get:

${r}^{4} = 4{r}^{2} ({ \cos^{2}\theta } - \sin^{2}\theta)$

Divide through by r²

${r}^{2} = 4 ({ \cos^{2}\theta } - \sin^{2}\theta)$

Apply the double angle identity

$\cos^{2}\theta -\sin^{2}\theta= \cos(2 \theta)$

The polar equation then becomes:

${r}^{2} = 4 \cos2\theta$

7 0

3 years ago

1. cot x sec4x = cot x + 2 tan x + tan3x

Mars2501 [29]

1. cot(x)sec⁴(x) = cot(x) + 2tan(x) + tan(3x)
cot(x)sec⁴(x) cot(x)sec⁴(x)
0 = cos⁴(x) + 2cos⁴(x)tan²(x) - cos⁴(x)tan⁴(x)
0 = cos⁴(x)[1] + cos⁴(x)[2tan²(x)] + cos⁴(x)[tan⁴(x)]
0 = cos⁴(x)[1 + 2tan²(x) + tan⁴(x)]
0 = cos⁴(x)[1 + tan²(x) + tan²(x) + tan⁴(4)]
0 = cos⁴(x)[1(1) + 1(tan²(x)) + tan²(x)(1) + tan²(x)(tan²(x)]
0 = cos⁴(x)[1(1 + tan²(x)) + tan²(x)(1 + tan²(x))]
0 = cos⁴(x)(1 + tan²(x))(1 + tan²(x))
0 = cos⁴(x)(1 + tan²(x))²
0 = cos⁴(x)        or         0 = (1 + tan²(x))²
⁴√0 = ⁴√cos⁴(x)    or   √0 = (√1 + tan²(x))²
0 = cos(x)   or   0 = 1 + tan²(x)
cos⁻¹(0) = cos⁻¹(cos(x)) or -1 = tan²(x)
90 = x       or        √-1 = √tan²(x)
i = tan(x)
(No Solution)

2. sin(x)[tan(x)cos(x) - cot(x)cos(x)] = 1 - 2cos²(x)
  sin(x)[sin(x) - cos(x)cot(x)] = 1 - cos²(x) - cos²(x)
sin(x)[sin(x)] - sin(x)[cos(x)cot(x)] = sin²(x) - cos²(x)
sin²(x) - cos²(x) = sin²(x) - cos²(x)
+ cos²(x) + cos²(x)
sin²(x) = sin²(x)
- sin²(x) - sin²(x)
0 = 0

3. 1 + sec²(x)sin²(x) = sec²(x)
sec²(x) sec²(x)
cos²(x) + sin²(x) = 1
  cos²(x) = 1 - sin²(x)
√cos²(x) = √(1 - sin²(x))
cos(x) = √(1 - sin²(x))
cos⁻¹(cos(x)) = cos⁻¹(√1 - sin²(x))
x = 0

4. -tan²(x) + sec²(x) = 1
-1   -1
  tan²(x) - sec²(x) = -1
tan²(x) = -1 + sec²
√tan²(x) = √(-1 + sec²(x))
tan(x) = √(-1 + sec²(x))
  tan⁻¹(tan(x)) = tan⁻¹(√(-1 + sec²(x))
x = 0

5 0

3 years ago