All categories

2 years ago

IMAGE BELOW I WILL GIVE BRAINLIEST

Mathematics

1 answer:

diamong [38]2 years ago

7 0

Answer:

for L 2, 2 and for LM it is -8,8

Step-by-step explanation:

easy

You might be interested in

Kite E F G H is inscribed in a rectangle. Points F and H are midpoints of sides of the rectangle, and creates a side length of x

Hoochie [10]

Answer:

5 units

Step-by-step explanation:

Let point O be the point of intersection of the kite diagonals.

|OF| = 2, |OH| = 5

|FH| = |OF| + |OH| = 2 + 5 = 7

FH and EG are the diagonals of the kite. Hence the area of thee kite is:

Area of kite EFGH = (FH * EG) / 2

Substituting:

35 = (7 * |EG|) / 2

|EG| * 7 = 70

|EG| = 10 units

The longer diagonal of a kite bisects the shorter one, therefore |GO| = |EO| = 10 / 2 = 5 units

x = |GO| = |EO| = 5 units

7 0

3 years ago

Read 2 more answers

An equation parallel and perpendicular to 4x+5y=19

mestny [16]

Answer:

An equation parallel to 4x + 5y = 19 would be y = -4/5x +12.

An equation perpendicular to 4x + 5y = 19 would be y = 5/4x + 10.

Step-by-step explanation:

The equation given represents a linear equation in Standard Form (Ax + By = C). Lines that are parallel to each other go the same direction and don't touch, so their slopes must be the same. However, lines that are perpendicular go in opposite directions and intersect, so their slopes must be the direct opposite of each other. In order to find the slope, you must first convert from the Standard Form given to Slope Intercept Form (y = mx +b). When you solve the given equation for 'y', you get: y = -4/5x + 19, where the slope = -4/5. To make a parallel equation, simply keep the same slope and choose a different y-intercept ('b'). To make a perpendicular equation, take the direct opposide of your slope 5/4 (positive) and choose a different y-intercept.

8 0

3 years ago

Find sinϴ and cosϴ if tanϴ=1/4 and sinϴ>0

eduard

If you're using the app, try seeing this answer through your browser: brainly.com/question/2787701

_______________

$\mathsf{tan\,\theta=\dfrac{1}{4}\qquad\qquad(sin\,\theta\ \textgreater \ 0)}\\\\\\
\mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{1}{4}}\\\\\\
\mathsf{4\,sin\,\theta=cos\,\theta\qquad\quad(i)}$

Square both sides:

$\mathsf{(4\,sin\,\theta)^2=(cos\,\theta)^2}\\\\
\mathsf{4^2\,sin^2\,\theta=cos^2\,\theta}\\\\
\mathsf{16\,sin^2\,\theta=cos^2\,\theta\qquad\qquad(but,~cos^2\,\theta=1-sin^2\,\theta)}\\\\
\mathsf{16\,sin^2\,\theta=1-sin^2\,\theta}$

$\mathsf{16\,sin^2\,\theta+sin^2\,\theta=1}\\\\
\mathsf{17\,sin^2\,\theta=1}\\\\
\mathsf{sin^2\,\theta=\dfrac{1}{17}}\\\\\\
\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{1}{17}}}\\\\\\
\mathsf{sin\,\theta=\pm\,\dfrac{1}{\sqrt{17}}}$

Since $\mathsf{sin\,\theta}$ is positive, you can discard the negative sign. So,

$\mathsf{sin\,\theta=\dfrac{1}{\sqrt{17}}\qquad\quad\checkmark}$

Substitute this value back into $\mathsf{(i)}$ to find $\mathsf{cos\,\theta:}$

$\mathsf{4\cdot \dfrac{1}{\sqrt{17}}=cos\,\theta}\\\\\\
\mathsf{cos\,\theta=\dfrac{4}{\sqrt{17}}\qquad\quad\checkmark}$

I hope this helps. =)

Tags: <em>trigonometric identity relation trig sine cosine tangent sin cos tan trigonometry precalculus</em>

7 0

3 years ago

Help me I need to get a 100

swat32

Answer:

option 2

Step-by-step explanation:

3 0

3 years ago

1. At a summer job, Melissa fixes computers for small businesses. She charges a $30 fee to start a job and $15 per hour for each

zalisa [80]

Thanks I 6889x the money 30x plus 1

7 0

3 years ago