All categories

2 years ago

What is the slope of a line that contains the ordered pairs (2,6) and (3,9)?

Mathematics

1 answer:

liubo4ka [24]2 years ago

4 0

Answer:

Step-by-step explanation:

→ Calculate the change in y coordinates

9 - 6 = 3

→ Calculate the change in x coordinates

3 - 2 = 1

→ Divide the answers

3 ÷ 1 = 3

You might be interested in

Male and female high school students reported how many hours they worked each week in summer jobs. The data is represented in th

KATRIN_1 [288]

Answer:

I'm 99% sure the answer is A

Step-by-step explanation:

8 0

3 years ago

Suppose angles A and B are complementary angles and m∠A = (x-5)° and m∠B = (2x+20)° Solve for X and then find m∠A and m∠B

MariettaO [177]

The angles are x =25, angle A = 20 and B = 70

<h3>How to solve for the angles?</h3>

The given parameters are:

A = x - 5

B = 2x + 20

Both angles are complementary angles.

This means that:

x - 5 + 2x + 20 = 90

Evaluate the like terms

3x = 75

Divide both sides by 3

x = 25

Substitute x = 25 in A = x - 5 and B = 2x + 20

A = 25 - 5 = 20

B = 2 * 25 + 20 = 70

Hence, the angles are x =25, angle A = 20 and B = 70

Read more about complementary angles at:

brainly.com/question/98924

#SPJ1

3 0

2 years ago

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

2 years ago

Passes through (-2,6) with a slop of -1/4

nikitadnepr [17]

Answer:

y=-1/4x-5 1/2.

Step-by-step explanation:

Using the formula y-y1=m(x-x1), plug in the slope and the numbers and you'll get y=-1/4x-5 1/2.

5 0

3 years ago

Read 2 more answers

Word problem:

Jlenok [28]

Answer:

Below.

Step-by-step explanation:

30 kg 500g + 28kg 700g

= 58kg + 1200g

= 59kg 200g.

Difference

= 30 kg 500g - 28kg 700g

= 2kg + 500 - 700

= 2kg - 200g

= 1kg 800g.

3 0

3 years ago