All categories

3 years ago

How are the fish food box and the shipping box similar and how are they different

Mathematics

1 answer:

GarryVolchara [31]3 years ago

3 0

They are similar because of the color and they are different by the size and shape hope that helps you

You might be interested in

Angle UZT is congruent to angle?

Ymorist [56]

Answer:

In order for me to help you I need to either see a photo or know more about the other angle.

Step-by-step explanation:

3 0

3 years ago

IDENTIFY LIKE TERMS<br> 5y + 15 - 80 - 10y

dexar [7]

Answer:

5y and -10y

15 and -80

-5y -65

3 0

2 years ago

Read 2 more answers

Find the general expression for the slope of a line tangent to the curve of y=2x^2+4x at the point P(x,y) . Then find the slopes

ArbitrLikvidat [17]

Complete Question

Find the general expression for the slope of a line tangent to the curve of y=2x^2+4x at the point P(x,y) . Then find the slopes for $x = 3$ and x=0.5. Sketch the curve and the tangent lines. What is the general expression for the slope of a line tangent to the curve of the function y=2x^2+4 at the point P(x,y) ?

Answer:

The generally expression for the slope of $y = 2x^2 + 4x$ is $y' = 4x +4$

The graph is shown on the first uploaded image

The generally expression for the slope of $y=2x^2+4$ is $y' = 4x$

Step-by-step explanation:

From the question we are told that

The equation of the curve is $y = 2x^2 + 4x$

First we differentiate the equation

$y' = 4x +4$

Therefore the generally expression for the slope tangent to the curve $y=2x^2+4x$ is $y' = 4x +4$

The next step is to substitute for x = 3 and x = 0.5

So for $x_1 = 3$

$y' = 4(3) +4$

$y' =m_1= 16$

And for $x_2 = 0.5$

$y' = 4(0.5) +4$

$y' =m_2= 6$

Here m_1 and m_2 are slops of the curve

Next we obtain the coordinates of the tangent lines

So at $x_1 = 3$

$y_1 = 2(3)^2 + 4(3)$

$y_1 = 21$

So the coordinate for the first tangent line is

$(x_1 , y_1 ) = (3 , 21)$

At $x_2 = 0.5$

$y_2 = 2(0.5)^2 + 4(0.5)$

=> $y_2 = 2.5$

So the coordinate for the second tangent line is

$(x_2 , y_2 ) = (0.5 , 2.5)$

Next we obtain the equation for the tangent lines

So generally the slope is mathematically represented as

$m = \frac{y - y_1 }{x-x_1}$

For $(x_1 , y_1 ) = (-3 , 21)$ and $y' =m_1= 16$

$16 = \frac{y -21 }{x-3)}$

=> $y = 16x - 27$

For $(x_2 , y_2 ) = (0.5 , 2.5)$ and $y' =m_2= 6$

$6 = \frac{y -2.5 }{x-0.5}$

$y = 6x -0.5$

Generally the general expression for the slope of a line tangent to the curve of the function y=2x^2+4 at the point P(x,y) is mathematically evaluated by differentiating y=2x^2+4 as follows

$y' = 4x$