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3 years ago

Which of the following ratios is used in finding the tangent of an angle in a right triangle?

1 answer:

8 0

Looking at the triangle from the aspect of the reference angle, the tangent of that angle would be the ratio of the side opposite the angle over the side adjacent to the angle (which is NOT the hypotenuse!)

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1. In the following diagram, UW bisects XV at Y. Which of the

aev [14]

Answer:

Statement 4 is correct

Step-by-step explanation:

Here, we want to select which statement is true based on the given diagram;

The statement that must be true is that Y is the midpoint of XV

This is because, by bisection , we mean dividing into 2 equal parts

The line UW has divided the line XV into two equal parts

So this mean that Y is the midpoint of the line XV

6 0

3 years ago

The distance from the tip of a slice of pizza to the crust is 7 inches.

Mice21 [21]

Answer:

radius

Step-by-step explanation:

4 0

2 years ago

Rita earns $17.00 per hour. If she gets a 6% raise, what will be her new hourly wage?

ipn [44]

multiply 17 by 6%

17 * 0.06 = 1.02 increase

17 +1.02 = 18.02 new hourly wage

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3 years ago

John went on a trip. 2 hours in he a 15 gallons of gas in her tank.7 hours into her trip she had 3 gallons left. Find the rate o

maks197457 [2]

Answer:

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5 0

3 years ago

Sketch the equilibrium solutions for the following DE and use them to determine the behavior of the solutions.

GREYUIT [131]

Answer:

$y=\dfrac{1}{1-Ke^{-t}}$

Step-by-step explanation:

Given

The given equation is a differential equation

$\dfrac{dy}{dt}=y-y^2$

$\dfrac{dy}{dt}=-(y^2-y)$

By separating variable

⇒ $\dfrac{dy}{(y^2-y)}=-t$

$\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-dt$

Now by taking integration both side

$\int\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-\int dt$

⇒ $\ln (y-1)-\ln y=-t+C$

Where C is the constant

$\ln \dfrac{y-1}{y}=-t+C$

$\dfrac{y-1}{y}=e^{-t+c}$

$\dfrac{y-1}{y}=Ke^{-t}$

$y=\dfrac{1}{1-Ke^{-t}}$

from above equation we can say that

When t will increases in positive direction then $e^{-t}$ will decreases it means that ${1-Ke^{-t}}$ will increases, so y will decreases. Similarly in the case of negative t.

4 0

3 years ago