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Anestetic [448]

2 years ago

12

Draw a line representing the “rise” and a line representing the “run” of the line. State the slope of the line in the simplest f

orm

1 answer:

irina1246 [14]2 years ago

4 0

Answer:

the rise would be the part of the line that is going up, and the run is the part of the line that is moving allong the x axis

Step-by-step explanation:

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KIM [24]

Answer:

a,c

Step-by-step explanation:

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3 years ago

Molodets [167]

Answer:

198.00

Step-by-step explanation:

I am not sure but I hope this helps some.

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3 years ago

Help I am very bad at math I need answers by tomorrow if you can help please do

d1i1m1o1n [39]

Answer:

Step-by-step explanation:

(a - b)(a +b) = a² - b²

1 - Sin² A = Cos² A

$LHS = \frac{1}{1- Sin A} + \frac{1}{1 + Sin A}\\\\= \frac{1*(1 + Sin A)}{(1- Sin A)(1 + Sin A)} + \frac{1*(1- Sin A)}{(1 + Sin A)(1- Sin A)}\\\\= \frac{1 + Sin A+ 1 - Sin A}{1^{2}- Sin^{2} A}\\\\= \frac{2}{1 - Sin^{2} A}\\\\= \frac{2}{Cos^{2} A}\\\\= 2 Sec^{2} A$

2) Sec² A - Tan² A = 1

$LHS = \frac{1}{Sec A - Tan A}\\\\=\frac{1*(Sec A + Tan A)}{(Sec A - Tan A)(Sec A + Tan A)}\\\\=\frac{Sec A + Tan A}{Sec^{2} A - Tan^{2} A}\\\\=\frac{Sec A + Tan A }{1}\\\\= Sec A + Tan A = RHS\\\\\\$

3) LHS = Cosec² A + Cot² A

= Cosec² A + Cosec² A - 1

= 2Cosec² A - 1 = RHS

$4) LHS = \frac{Sec A}{Cos A}- \frac{Tan A}{Cot A}\\\\ = Sec A*\frac{1}{Cos A}-Tan A*\frac{1}{Cot A}\\\\ = Sec A * Sec A - Tan A * Tan A\\\\= Sec^{2} A - Tan^{2} A \\\\= 1$

3 0

3 years ago

Solve the inequality <br> -5x > 35

galina1969 [7]

-5x > 35
-x > 7
(Divide it by 5)

5 0

3 years ago

Read 2 more answers

A cone has a radius of 5 ft and a height of 15 ft. It is empty and is being filled with water at a constant rate of 15 15 ft 3 f

kvasek [131]

Answer:

$\frac{dr}{dt}=0.11062ft/sec$

Step-by-step explanation:

From the question we are told that:

Radius $r=5ft$

Height $H=15ft$

Rate $R=15ft/3sec =5ft/s$

Surface Radius $R_{surf}=2.2ft$

Generally the equation for Volume is mathematically given by

$V=\frac{1}{3}\pi*r^2h$

Since radius to height ratio gives

$\frac{r}{h}=\frac{5}{15}$

$\frac{r}{h}=\frac{1}{3}$

$h=3r$

Therefore

$V=\frac{1}{3}\pi*r^2(3r)$

$V=\pi r^3$

Generally the equation for Change of Volume is mathematically given by

$\frac{dv}{dt}=\pi \frac{d}{dt}(r^3)$

$\frac{dv}{dt}=\pi 3*r^2 \frac{dr}{dt}$

$\frac{dv}{dt}=\pi 3*(2.2)^2 \frac{dr}{dt}$

$\frac{dr}{dt}=\frac{5}{45.62}$

$\frac{dr}{dt}=0.11062ft/sec$

3 0

3 years ago