<h3><u><em>Option B</em></u></h3><h3><u><em>The equation of a line, in point- slope form, that passes through (5, -3) and has a slope of 2/3</em></u></h3><h3><u><em /></u>
![y + 3 = \frac{2}{3}(x -5)](https://tex.z-dn.net/?f=y%20%2B%203%20%3D%20%5Cfrac%7B2%7D%7B3%7D%28x%20-5%29)
<u><em /></u></h3>
<em><u>Solution:</u></em>
<em><u>The equation of line in point slope form is given as:</u></em>
![y - y_1 = m(x-x_1)](https://tex.z-dn.net/?f=y%20-%20y_1%20%3D%20m%28x-x_1%29)
Where, "m" is the slope of line
Given that,
slope = 2/3
![m = \frac{2}{3}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B2%7D%7B3%7D)
The line passes through (5, -3)
Substitute m = 2/3 and
= (5, -3) in point slope form
![y - (-3) = \frac{2}{3}(x -5)\\\\y + 3 = \frac{2}{3}(x -5)](https://tex.z-dn.net/?f=y%20-%20%28-3%29%20%3D%20%5Cfrac%7B2%7D%7B3%7D%28x%20-5%29%5C%5C%5C%5Cy%20%2B%203%20%3D%20%5Cfrac%7B2%7D%7B3%7D%28x%20-5%29)
Thus equation of line in point slope form is found
Answer:
<u>y = w and ΔABC ~ ΔCDE</u>
Step-by-step explanation:
Given sin(y°) = cos(x°)
So, ∠y + ∠x = 90° ⇒(1)
And as shown at the graph:
ΔABC is aright triangle at B
So, ∠y + ∠z = 90° ⇒(2)
From (1) and (2)
<u>∴ ∠x = ∠z </u>
ΔCDE is aright triangle at D
So, ∠x + ∠w = 90° ⇒(3)
From (1) and (3)
<u>∴ ∠y = ∠w</u>
So, for the triangles ΔABC and ΔCDE
- ∠A = ∠C ⇒ proved by ∠y = ∠w
- ∠B = ∠D ⇒ Given ∠B and ∠D are right angles.
- ∠C = ∠E ⇒ proved by ∠x = ∠z
So, from the previous ΔABC ~ ΔCDE by AAA postulate.
So, the answer is <u>y = w and ΔABC ~ ΔCDE</u>
Answer:
b = 29
Step-by-step explanation:
C x D = A x B
15(11x - 91) = 7(2x + 10)
165x - 1365 = 14x + 70
add 1365 to both sides of the equation:
165x = 14x + 1435
subtract 14x from both sides:
151x = 1435
divide both sides by 151:
x = 9.5
b = 2(9.5) + 10 = 29
Answer: 180 degrees
Step-by-step explanation:
A rotation of 180 degrees about the origin maps
.