All categories

3 years ago

Help me ASAP please I don't understand it please help It is dued by TONIGHT

Mathematics

1 answer:

Sergeu [11.5K]3 years ago

3 0

So, the first blank is for what kind of angle it is: acute, obtuse, reflex, straight, or right. The second blank is for what the measure of the angle is. You can find that out by looking at the protractor. It should tell you what the measure of the angle is. You can read the protractor by looking at the numbers. The numbers are the degrees of incline the angle measures, which is what you're looking for.

You might be interested in

Five friends share 3 bags of trail mix equally. What fraction of a bag does EA friend get

Vesna [10]

I did this on a calcalator 3÷5=0.6

7 0

3 years ago

Read 2 more answers

4, 3, 2, 5, 6, 6, 10, 5, 6, 2, 3, 4, 6, 7, 14,5<br><br> 3. What is the mode(s) of the data set?

guapka [62]

Answer:

Step-by-step explanation:

6 shows up 4 times

5 0

3 years ago

Read 2 more answers

The cost to mail a letter is a base charge of $0.42 plus $0.17 for each ounce. Write an equation that could be used to find how

tresset_1 [31]

Answer:

D) 0.42 + 0.17 x = 0.93

Step-by-step explanation:

0.93 - 0.42 = 0.51

0.51 / 0.17 = 3

3 ounces and the answer is D

3 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.