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3 years ago

What is the horizontal distance between (−5/3, -2) and (4, -2)?

Mathematics

1 answer:

Nataly [62]3 years ago

6 0

Answer:

-2?

Step-by-step explanation:

use y2-y1 over x2-x1 to find the answer. So you would do -2--5/3 over 4--2. The two negitives turn into a positive. SO it is now -2+5/3 over 4+2. -2+5/3= -1/3 over 4+2=6 it is now -1/3/6 which equals -2. So I think the answer is -2.<u> I do not know if you should trust me on this but I tried. </u>

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The triangles are similar by:

Tamiku [17]

Answer:

E. by the SAS similarity theorem.

Step-by-step explanation:

Included angle x° in ∆ ABC ≅ included angle x° in ∆EDC (vertical angles are equal)

DC/BC = 240/150 = 1.6

EC/AC = 320/200 = 1.6

This implies that the ratio of two corresponding sides of both triangles are the same.

Two triangles are considered similar to each other by the SAS similarity theorem of they have a corresponding included angle that is equal and two corresponding sides that are congruent to each other. Therefore, both triangles are similar by the SAS similarity theorem.

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3 years ago

A candy company puts 3 gumdrops in each bag. How many gumdrops will the company need to fill 5,000 bags?

Ksenya-84 [330]

15,000
3 multiplied by 5,000 is 15,000

5 0

1 year ago

Which equation has the same solution as 10x-x+5=4110 x − x + 5 = 41?

Tatiana [17]

D or the last dot how ever you wanna describe it

6 0

3 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

1 year ago

Suppose you work as a clerk at a local retail store earning $6 per hour. The amount

rusak2 [61]

Answer:

the independent variable is amount being paid per hour, as the final outcomes relys on that answer. the dependent variable is the hours worked, as thats the variable that can be changed.

7 0

2 years ago