Ask question

Ask question

All categories

3 years ago

8

Select the correct answer. If the function 5x + y = 1 has the domain (-2, 1, 6), then what is the corresponding range? OA. {-9,

6, 31) log OB. {9, -6, -31} Oc. {-11, 4, 29) OD. {11,-4,-29}

1 answer:

Leokris [45]3 years ago

3 0

Answer:

D) {11,-4,-29}

Step-by-step explanation:

rewrite the equation as y= -5x+1 and plug in -2,1 and 6 to get y

-5*-2+1=11

-5*1+1=-4

-5*6+1=-29

You might be interested in

Starting at the park, Jake bikes 3 blocks north, 4 blocks east, 4 blocks south, 7 blocks west, and 1 block north. How many block

yan [13]

I'll use a coordinate plane for this since it's easier (for me) :)

Red = park
Blue = his trail
Pink = distance from park = 3

Hope this helped! Please comment or DM me if you have anymore questions or don't understand :)

7 0

3 years ago

What is the measure of angle C?

Aleksandr-060686 [28]

Answer:

38 or 76 but i think its 76

Step-by-step explanation:

3 0

3 years ago

Water is added to a cylindrical tank of radius 5 m and height of 10 m at a rate of 100 L/min. Find the rate of change of the wat

nirvana33 [79]

Answer:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

$\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}$

And solving for $\frac{dh}{dt}$ we got:

$\frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}$

We need to convert the rate given into m^3/min and we got:

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

5 0

3 years ago

Using either the special 45-45 right triangle or the Unit circle determine what the exact value

WITCHER [35]

Positive root 2/2 (the last one)

6 0

3 years ago

Which number is the answer

const2013 [10]

Answer:

1

Step-by-step explanation:

32-24 gives you 8

So she only needs one.

7 0

3 years ago