Which similarity postulate or theorem let’s you conclude that JKL = MNO?

Which scatter plot best represents a linear association between y and x ?

dedylja [7]

Answer:

D

Step-by-step explanation:

it looks more like a line.

A is exponential

B is a parabola

C is no association

6 0

2 years ago

In a bag, there are three red marbles, four blue marbles, two yellow marbles and three black marbles. What is the probability of

charle [14.2K]

Answer:

3/12 chance

Step-by-step explanation:

3 0

3 years ago

What is the measure of ZB? 28 36 B [?] degrees Enter

Sindrei [870]

All triangles have a measurement of 180 degrees. Therefore, we can use this information to make an equation to find the value of angle B.

Angle B = 180 - (28 + 36)
Angle B = 180 - 64
Angle B = 116

So, angle B is equal to 116 degrees.

4 0

3 years ago

Identify sinJ as a fraction and as a decimal rounded to the nearest hundredth.

KiRa [710]

In the given right triangle, the trigonometric ratio, sin J has a fractional value of 5/13 and a decimal value of 0.39.

In trigonometry, for a right triangle, the sine (sin) of any angle θ is given as the ratio of its opposite side to the hypotenuse of the triangle, that is, sin θ = (opposite side)/(hypotenuse).

In the question, we are asked to find the trigonometric ratio, sin J, for the given right triangle JKL.

The side opposite to angle J is KL, which has a value of 3 units.

The hypotenuse of the given right triangle is JK, which has a value of 7.8 units.

Thus, sin J can be calculated as:

sin J = KL/JK = 3/7.8 = 5/13 = 0.39.

Thus, in the given right triangle, the trigonometric ratio, sin J has a fractional value of 5/13 and a decimal value of 0.39.

Learn more about trigonometric ratios at

brainly.com/question/20367642

#SPJ1

8 0

1 year ago

Simplify . (1/c + 1/h)/(1/(c ^ 2) - 1/(r ^ 2))

Marrrta [24]

Answer:

$\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

Step-by-step explanation:

$\frac{\frac{1}{c}+\frac{1}{h}}{\frac{1}{c^2}-\frac{1}{r^2}}$

Combine $\frac{1}{c} + \frac{1}{h}$

$\frac{\frac{h+c}{ch}}{\frac{1}{c^2}-\frac{1}{r^2}}$

Combine the bottom, too.

$=\frac{\frac{h+c}{ch}}{\frac{r^2-c^2}{c^2r^2}}$

Apply the fraction rule

$=\frac{\left(h+c\right)c^2r^2}{ch\left(r^2-c^2\right)}$

Cancel

$=\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

Therefore, $\frac{\left(\frac{1}{c}+\frac{1}{h}\right)}{\left(\frac{1}{\left(c^2\right)}-\frac{1}{\left(r^2\right)}\right)}:\quad \frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

5 0

3 years ago