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3 years ago

The point (- 3, 2) is on the line given by which equation below?

Mathematics

2 answers:

blondinia [14]3 years ago

7 0

What equation below, there is no information below

My name is Ann [436]3 years ago

5 0

Answer:

what equation below? :/

Step-by-step explanation:

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Four students were asked to write an expression which has terms that have a greatest common factor of 8c. The expressions provid

wel

Answer:

Wyatt wrote the correct expression

Step-by-step explanation:

Here we are to evaluate which of the students wrote an expression with the highest common factor being 8c

For Tanya; 32c + 16c; 16c( 2 + 1); This is wrong as the highest common factor is 16c

For Wyatt, we have 8c - 48cd; factorizing, we have 8c(1-6cd); This is correct

For Xavier we have 36c-24c; 8 is not a factor of 36, so it won’t work

For Yang 8c + 40; 8(c+ 5); what we are trying to look for is 8c and not 8 so this is wrong also

3 0

3 years ago

Read 2 more answers

an inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a

viktelen [127]

Answer:

the rate of change of the water depth when the water depth is 10 ft is; $\mathbf{\dfrac{dh}{dt} = \dfrac{-25}{100 \pi} \ \ ft/s}$

Step-by-step explanation:

Given that:

the inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec.

We are meant to find the rate of change of the water depth when the water depth is 10 ft.

The diagrammatic expression below clearly interprets the question.

From the image below, assuming h = the depth of the tank at a time t and r = radius of the cone shaped at a time t

Then the similar triangles ΔOCD and ΔOAB is as follows:

$\dfrac{h}{r}= \dfrac{20}{8}$ ( similar triangle property)

$\dfrac{h}{r}= \dfrac{5}{2}$

$\dfrac{h}{r}= 2.5$

h = 2.5r

$r = \dfrac{h}{2.5}$

The volume of the water in the tank is represented by the equation:

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

$V = \dfrac{1}{3} \pi (\dfrac{h^2}{6.25}) h$

$V = \dfrac{1}{18.75} \pi \ h^3$

The rate of change of the water depth is :

$\dfrac{dv}{dt}= \dfrac{\pi r^2}{6.25}\ \dfrac{dh}{dt}$

Since the water is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec

Then,

$\dfrac{dv}{dt}= - 4 \ ft^3/sec$

Therefore,

$-4 = \dfrac{\pi r^2}{6.25}\ \dfrac{dh}{dt}$

the rate of change of the water at depth h = 10 ft is:

$-4 = \dfrac{ 100 \ \pi }{6.25}\ \dfrac{dh}{dt}$

$100 \pi \dfrac{dh}{dt} = -4 \times 6.25$

$100 \pi \dfrac{dh}{dt} = -25$

$\dfrac{dh}{dt} = \dfrac{-25}{100 \pi}$

Thus, the rate of change of the water depth when the water depth is 10 ft is; $\mathtt{\dfrac{dh}{dt} = \dfrac{-25}{100 \pi} \ \ ft/s}$

4 0

3 years ago

Two mechanics worked on a car. The first mechanic worked for 20 hours, and the second mechanic worked for 5 hours. Together they

Vesna [10]

Answer:

Step-by-step explanation:

7 0

3 years ago

3+3+3+3+3 what is the product?

lukranit [14]

The correct answer is 15

7 0

3 years ago

Read 2 more answers

E^(8x)+e^(-x) find the interval on which f is increasing

saveliy_v [14]

Assuming ya mean
$f(x)=e^{8x}+e^{-x}$
take derivitive
remember dy/dx e^x=e^x
also the chain rule

so
$f'(x)=8e^{8x}+-1e^{-x}$
find where it equals 0
$0=8e^{8x}+-e^{-x}$
$e^{-x}=8e^{8x}$
multiply both sides by e^x
$1=8e^{9x}$
divide both sides by 8
$\frac{1}{8}=e^{9x}$
take ln of both sides
$ln(\frac{1}{8})=ln(e^{9x})$
$ln(\frac{1}{8})=9x$
divide both sides by 9
$\frac{ln(\frac{1}{8})}{9}=x$

x≈-1/5
f'(-1)<0
f'(0)>0

so it is increasing from x to infinity

the interval it is increasing on is $(\frac{ln(\frac{1}{8})}{9}, \infty)$

6 0

2 years ago