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3 years ago

Find the value of \mathrm{x}xin the following figure.

Mathematics

1 answer:

Hoochie [10]3 years ago

3 0

Step-by-step explanation:

step 1. I guess we assumed the two lines across the transversal are parallel

step 2. 3x + 1 = 85 (definition of alternate exterior angles)

step 3. 3x = 84 (subtract 1 from each side)

step 4. x = 28. (divide both sides by 3)

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Write an equation of the line in slope-intercept from?

netineya [11]

Answer:

$y=-\frac{3}{2}x+2$

Step-by-step explanation:

This is for the first question.

Slope-intercept is:

$y=mx+b$

m - slope

b - y-intercept

We are given the slope of $-\frac{3}{2}$ and a y-intercept of 2.

Plug in the values:

$y=mx+b\rightarrow\boxed{y=-\frac{3}{2}x+2}$

Hope this helps.

4 0

3 years ago

The degree of the function f(x) = -(x + 1)2(2x - 3)(x + 2)2 is , and its y-intercept is

Serggg [28]

Answer:

Degree = 5

Y intercept = 12

Step-by-step explanation:

our function is given as

$f(x)=-(x+1)^2 \cdot (2x-3) \cdot (x+2)^2$

in the above polynomial our variable is x and it is being multiplied for 5 times. Hence the probable degree for the above polynomial is 5

In order to find the y intercept , we need to put x=0 in f(x) as y intercept is the point at which the function graphically meets y axis and where x = 0

Hence

$f(0)=-(0+1)^2 \cdot (2(0)-3) \cdot (0+2)^2$

$f(0)=-(1)^2 \cdot (0-3) \cdot (2)^2$

$f(0)=-1 \cdot (-3) \cdot 4$

$f(0)=3 \cdot 4$

$f(0)=3 \times 4$

$f(0)=12$

Hence the y intercept is 12 units

6 0

2 years ago

At the market 7 apples cost 4.55 how much would 9 apples cost

I am Lyosha [343]

Your answer is 22.09. I hope you get right

3 0

2 years ago

ILL MARK YOU BRAINLIEST Factor completely 3ab(x + 1) − 2(x + 1).

Nuetrik [128]

Part one of answering how to factor

3 0

2 years ago

How to solve this trig

n200080 [17]

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

First Question

What we have to do is to isolate cos first.

$\displaystyle \large{ cos \theta = - \frac{1}{2} }$

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

Find Q2

$\displaystyle \large{ \pi - \frac{ \pi}{3} = \frac{3 \pi}{3} - \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{2 \pi}{3} }$

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{3} = \frac{3 \pi}{3} + \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{4 \pi}{3} }$

Both values are apart of the interval. Hence,

$\displaystyle \large \boxed{ \theta = \frac{2 \pi}{3} , \frac{4 \pi}{3} }$

Second Question

Isolate sin(4 theta).

$\displaystyle \large{sin 4 \theta = - \frac{1}{ \sqrt{2} } }$

Rationalize the denominator.

$\displaystyle \large{sin4 \theta = - \frac{ \sqrt{2} }{2} }$

The problem here is 4 beside theta. What we are going to do is to expand the interval.

$\displaystyle \large{0 \leqslant \theta < 2 \pi}$

Multiply whole by 4.

$\displaystyle \large{0 \times 4 \leqslant \theta \times 4 < 2 \pi \times 4} \\ \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}$

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{4} = \frac{ 4 \pi}{4} + \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{5 \pi}{4} }$

Find Q4

$\displaystyle \large{2 \pi - \frac{ \pi}{4} = \frac{8 \pi}{4} - \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{7 \pi}{4} }$

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

$\displaystyle \large{ \theta + 2 \pi k = \theta \: \: \: \: \: \sf{(k \: \: is \: \: integer)}}$

Hence:-

For Q3

$\displaystyle \large{ \frac{5 \pi}{4} + 2 \pi = \frac{13 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 4\pi = \frac{21 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 6\pi = \frac{29 \pi}{4} }$

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

For Q4

$\displaystyle \large{ \frac{ 7 \pi}{4} + 2 \pi = \frac{15 \pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 4 \pi = \frac{23\pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 6 \pi = \frac{31 \pi}{4} }$

Therefore:-

$\displaystyle \large{4 \theta = \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4} }$

Then we divide all these values by 4.

$\displaystyle \large \boxed{\theta = \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16} }$

Let me know if you have any questions!

3 0

2 years ago