All categories

3 years ago

For y=3x+5, find the rangevalue the domain value is 5.I NEED IT ASAP

Mathematics

2 answers:

liubo4ka [24]3 years ago

6 0

Answer: 20

Step-by-step explanation: Have a great night

bogdanovich [222]3 years ago

4 0

Answer:

y = 20

Step-by-step explanation:

y = 3x + 5 If x = 5

y = (3*5) + 5

y = 15 + 5

y = 20

You might be interested in

What are the domain restrictions of the expression?

lubasha [3.4K]

h^2 - 12h + 20 can be factored to:

(h-10)(h-2)

These cannot be zero so, h cannot be 10 or 2.

7 0

3 years ago

Read 2 more answers

-17x + 2 +3x^2 + 2x^3 divided by x-2

Stells [14]

Answer:

2x^2+7x-3

remainder is -4

Step-by-step explanation:

Check the image

6 0

3 years ago

Carlos find the absolute value of -5 3/10 and then find is the opposite of his answer Jason finds the opposite of -5 3/10 and th

a_sh-v [17]

we know that

$5 \frac{3}{10}= \frac{5*10+3}{10} = \frac{53}{10}$

Step 1

Carlos's method

1) Find the absolute value of $-\frac{53}{10}$

$\left|-\frac{53}{10}\right|=\frac{53}{10}$

2) Find the opposite

$opposite=-\frac{53}{10}$

Step 2

Jason's method

1) Find the opposite of $-\frac{53}{10}$

$opposite=\frac{53}{10}$

2) Find the absolute value

$\left|\frac{53}{10}\right|=\frac{53}{10}$

Step 3

Compare the answers

$\frac{53}{10} > -\frac{53}{10}$

therefore

The answer is

The greater value is obtained with the method of Jason

4 0

3 years ago

Which of the following equations will yield the volume of a triangular prism?

Flura [38]

Answer:

V = Bh

Step-by-step explanation:

The equations will yield the volume of a triangular prism is V = Bh.

Where,

V = Volume

B = Base

h = Height

Here are an example :

brainly.com/question/17421674

brainly.com/question/16420628

For more information:

brainly.com/question/11508323

Kavinsky

5 0

2 years ago

Use chain rule to find dw/dt w=ln(x^2 + y^2 + z^2)^(1/2) , x=sint, y=cost, z=tant

son4ous [18]

The chain rule states that

$\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\partial w}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial w}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac{\partial w}{\partial z}\dfrac{\mathrm dz}{\mathrm dt}$

Since $\ln(x^2+y^2+z^2)^{1/2}=\dfrac12\ln(x^2+y^2+z^2)$ , you have

$\dfrac{\partial w}{\partial x}=\dfrac x{x^2+y^2+z^2}$
$\dfrac{\partial w}{\partial x}=\dfrac y{x^2+y^2+z^2}$
$\dfrac{\partial w}{\partial z}=\dfrac z{x^2+y^2+z^2}$

and

$\dfrac{\mathrm dx}{\mathrm dt}=\cos t$
$\dfrac{\mathrm dy}{\mathrm dt}=-\sin t$
$\dfrac{\mathrm dz}{\mathrm dt}=\sec^2t$

Also, since $x^2+y^2+z^2=\sin^2t+\cos^2t+\tan^2t=1+\tan^2t=\sec^2t$ , the derivative is

$\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\sin t\cos t}{\sec^2t}-\dfrac{\sin t\cos t}{\sec^2t}+\dfrac{\tan t\sec^2t}{\sec^2t}$
$\dfrac{\mathrm dw}{\mathrm dt}=\tan t$

4 0

3 years ago