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3 years ago

Find the gradient of each of these line y = 2x + 4

2 answers:

8 0

We Know, Principle equation of a line is: y = mx + c
Here, we have: y = 2x + 4

Compare the equations,
Gradient of line (m) would be equal to 2

In short, Your Answer would be 2

Hope this helps!

Kitty [74]3 years ago

3 0

Y = 2x + 4

comparing to y = mx + c, where m = slope or gradient of the line and c = vertical intercept.

y = 2x + 4 and y = mx + c

m = 2

Hence the slope or gradient of the line = 2

Hope this explains it.

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Which graphs show continuous data?<br><br> Select each correct answer.

love history [14]

C

Due to the fact that it is starting from 0 and moving in a steady pattern

5 0

3 years ago

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Grace, Chelsea, and Roan are simplifying the same polynomial expression. which students work is correct and why?

Trava [24]

Answer:

Grace

Step-by-step explanation:

Chelsea and Roan didn't distribute the negative sign in the second half of the first expression.

-2(x - 8) is

-2(x) + (-2)(-8)

-2x + 16

8 0

3 years ago

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Write 0.125 as if it was 2 to a power

Westkost [7]

Answer:

$2^{-3}$

Step-by-step explanation:

0.125 as a fraction is 125/1000 or 1/8

1 is the same as $2^{0}$ (anything to the power of 0 is 1)
8 is the same as 2³ (2 × 2 × 2 = 8)
$2^{0}$ ÷ 2³ = $2^{-3}$ (You subtract the powers: 0 - 3 = -3)

Hope this helps!

4 0

3 years ago

(1 point) In this problem we show that the function f(x,y)=7x−yx+y f(x,y)=7x−yx+y does not have a limit as (x,y)→(0,0)(x,y)→(0,0

polet [3.4K]

Answer:

Step-by-step explanation:

Given that,

f(x, y)=7x−yx+y

We want to show that the limit doesn't exist as (x, y)→(0,0).

Limits typically fail to exist for one of four reasons:

1. The one-sided limits are not equal

2. The function doesn't approach a finite value

3. The function doesn't approach a particular value

4. The x - value is approaching the endpoint of a closed interval

a. Considering the case that y=3x

lim(x,y)→(0,0) 7x−yx+y

Since y=3x

lim(x,3x)→(0,0) 7x−3x(x)+3x

lim(x,3x)→(0,0) 10x−3x²

Therefore,

lim(x,3x)→(0,0) 10x−3x² = 0-0=0

b. Let also consider at y=4x

lim(x,y)→(0,0) 7x−yx+y

Since y=4x

lim(x,4x)→(0,0) 7x−4x(x)+4x

lim(x,4x)→(0,0) 11x−4x²

Therefore,

lim(x,4x)→(0,0) 11x−4x² = 0-0=0

c. Let also consider it generally at y=mx

lim(x,y)→(0,0) 7x−yx+y

Since y=mx

lim(x,mx)→(0,0) 7x−mx(x)+mx

lim(x, mx)→(0,0) (7+m)x−mx²

Therefore,

lim(x, mx)→(0,0) (7+m)x−mx² = 0-0=0

But the limit of the given function exist.

So let me assume the function is wrong and the question meant.

f(x, y)= (7x−y) / (x+y)

So, let analyze again

a. Considering the case that y=3x

lim(x,y)→(0,0) (7x−y)/(x+y)

Since y=3x

lim(x,3x)→(0,0) (7x−3x)/(x+3x)

lim(x,3x)→(0,0) 4x/4x

lim(x,3x)→(0,0) 1

Therefore,

lim(x,3x)→(0,0) 1= 1

So the limit is 1

b. Let also consider at y=4x

lim(x,y)→(0,0) (7x−y)/(x+y)

Since y=4x

lim(x,4x)→(0,0) (7x−4x)/(x+4x)

lim(x,4x)→(0,0) 3x/5x

lim(x,4x)→(0,0) 3/5

Therefore,

lim(x,4x)→(0,0) 3/5 = 3/5

So the limit is 3/5

This show that the limit does not exit.

Since one of the condition given above is met, then the limit does not exist. i.e. The function doesn't approach a particular value

c. Let also consider it generally at y=mx

lim(x,y)→(0,0) (7x−y)/(x+y)

Since y=mx

lim(x,mx)→(0,0) (7x−mx)/(x+mx)

lim(x,mx)→(0,0) (7-m)x/(1+m)x

lim(x, mx)→(0,0) (7-m)/(1+m)

Therefore,

lim(x, mx)→(0,0) (7-m)/(1+m) = (7m)/(1+m)

Then, the limit is (7-m)/(1+m)

So the limit doesn't not have a specific value, it depends on the value of m, so the limit doesn't exist.

7 0

3 years ago

What is the equation of the circle with center (4, 4) that passes through the point (10, 14)? HELP MEEEEE PLEASE

Gemiola [76]

Answer:

The equation of the circle can be written as:

$\left(x-4\right)^2+\left(y-4\right)^2=136$

Step-by-step explanation:

The general equation of a circle with center $(h,k)$ and radius $r$ is:

$\left(x-h\right)^2+\left(y-k\right)^2=r^2$

In our example, we know $\left(h,k\right)=\left(4,4\right)$ , as we just have to make sure we need determine $\:r^2$ .

$\left(x-4\right)^2+\left(y-4\right)^2=r^2\:\:$

As the circle passes through (10, 14), that pair of values for x and y must satisfy the equation. So we have:

$\left(10-4\right)^2+\left(14-4\right)^2=r^2$

$\mathrm{Switch\:sides}$

$r^2=\left(10-4\right)^2+\left(14-4\right)^2$

$r^2=6^2+10^2$

$r^2=36+100$

$r^2=136$

Thus the equation of the circle can be written as:

$\left(x-4\right)^2+\left(y-4\right)^2=136$

5 0

3 years ago

Read 2 more answers