All categories

3 years ago

Simplify: 14 - ³√216 No files

Mathematics

1 answer:

borishaifa [10]3 years ago

4 0

Answer:

Step-by-step explanation:

14 - ∛216

14 - 6

<em>Step 2: Subtract 6 from 14 to get 8.</em>

14 - ∛216 simplifies to 8.

You might be interested in

PLEASE HELPBFHDDHHDDHHD

Ierofanga [76]

Answer:what am I doing

Step-by-step explanation:

3 0

3 years ago

A rectangular picture has dimensions of 8 in by 12 in this picture is placed in a picture frame with thick border to Listen to B

rosijanka [135]

Answer: 6

Step-by-step explanation:

4 0

3 years ago

Which of the following is true? |−8| < 6 |−6| < |−8| |8| < |−6| |−6| < −8

tamaranim1 [39]

Answer: The second choice

Step-by-step explanation:

Absolute value is the distance a number is from 0 on the number line. Thus, the absolute value of -8 is 8, which is greater than the absolute value of -6, or 6.

<em>Hope it helps <3</em>

8 0

3 years ago

Read 2 more answers

Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the pa

konstantin123 [22]

Answer:

$y=2x-8$

Step-by-step explanation:

The given parametric equation is;

$x=6+ln(t),y=t^2+3$

<h3><u>BY ELIMINATING THE PARAMETER</u></h3>

To eliminate the parameter we make $t$ the subject in one equation and put it inside the other.

We make $t$ the subject in $x=6+ln(t)$ because it is easier.

$\Rightarrow x-6=ln(t)$

$\Rightarrow {e}^{x-6}=e^{ln(t)}$

$\Rightarrow {e}^{x-6}=t$

$t={e}^{x-6}$

We now substitute this into $y=t^2+3$ .

This gives us;

$y=(e^{x-6})^2+3$ .

$\Rightarrow y=e^{2(x-6)}+3$ .

We have now eliminated the parameter.

The equation of the tangent at (6,4) is given by;

$y-y_1=m(x-x_1)$

where the gradient function is given by;

$\frac{dy}{dx}=2e^{2(x-6)}$

We substitute $x=6$ into the gradient function to obtain the gradient.

$\Rightarrow m=2e^{2(6-6)}$

$\Rightarrow m=2e^0$

$\Rightarrow m=2$

The equation of the tangent becomes

$y-4=2(x-6)$

We simplify to obtain

$y=2x-12+4$

$y=2x-8$

<h3><u>WITHOUT ELIMINATING THE PARAMETER</u></h3>

The given parametric equation is;

$x=6+ln(t),y=t^2+3$

For $x=6+ln(t)$

$\frac{dx}{dt}=\frac{1}{t}$

For $y=t^2+3$

$\frac{dy}{dt}=2t$

The slope is given by;

$\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\frac{dy}{dx}=\frac{2t }{\frac{1}{t} }$

$\frac{dy}{dx}=2t^2$

At the point, (6,4), we plug in any of the values into the parametric equation and find the corresponding value for $t$ .

Notice that

When $x=6$ , $6=6+\ln(t)$

$6-6=\ln(t)$

$0=\ln(t)$

$e^0=e^\ln(t)$

$1=t$

when $y=4$ , $4=t^2+3$

$4-3=t^2$

$1=t^2$

$t=\pm1$

But the slope is the same when we plug in any of these values for t.

$\frac{dy}{dx}=2(\pm1)^2=2$

The equation of the tangent becomes

$y-4=2(x-6)$

We simplify to obtain

$y=2x-12+4$

$y=2x-8$

7 0

3 years ago

Find the value of x<br>Thanks!

statuscvo [17]

Hi there!

From the given information, we can infer that the inner quadrilateral has equal side lengths of x and 2x-3.

We can turn this into a mathematical equation with 1 variable.

x = 2x - 3

Subtract both sides by 2x

-x = -3

Multiply both sides by -1

x = 3

Have an awesome day! :)

~collinjun0827, Junior Moderator

7 0

3 years ago