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3 years ago

How to find the steeper slope in a graph?

Mathematics

2 answers:

Fed [463]3 years ago

6 0

Answer:

Step-by-step explanation:

The higher the y value is compared to the x value, the steeper the slope.

so, If we had two lines one with a slope of 5/1 and another with a slope of 4/1

Then the line with a slope of 5/1 would be greater.

I hope this wasn't confusing and hope this helps:)

Sauron [17]3 years ago

5 0

Answer:

When you look at the two lines, you can see that the blue line is steeper than the red line. It makes sense the value of the slope of the blue line, 4, is greater than the value of the slope of the red line, . The greater the slope, the steeper the line.

Step-by-step explanation:

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What is the value of the missing angle?a, 720b, 120c, 128d, 138

Zigmanuir [339]

The sum of all the interior angles in a hexagon is 720°, so if you add up all the angles and then take them away from 720, you'll get your answer:
152 + 85 = 237
237 + 85 = 322
322 + 125 = 447
447 + 135 = 582
So 720 - 582 = 138°, which is your answer. I hope this helps!

3 0

3 years ago

How do you graph -1/5x-1 and 4/5x-6

kakasveta [241]

Answer:

The points for the given to linear equations is (5 , - 2) and (5 , - 1)

The points is plotted on the graph shown .

Step-by-step explanation:

Given as :

The two linear equation are

y = $\dfrac{-1}{5}$ x - 1 ...........1

y = $\dfrac{4}{5}$ x - 6 ...........2

Now, Solving both the linear equations

Put the value of y from eq 2 into eq 1

I.e $\dfrac{4}{5}$ x - 6 = $\dfrac{-1}{5}$ x - 1

Or, $\dfrac{4}{5}$ x + $\dfrac{1}{5}$ x = 6 - 1

Or, $\dfrac{4 + 1}{5}$ x = 5

or, $\dfrac{5}{5}$ x = 5

∴ x = 5

Now, Put the value of x in eq 1

So, y = $\dfrac{-1}{5}$ x - 1

Or, y = $\dfrac{-1}{5}$ × 5 - 1

or, y = $\dfrac{-5}{5}$ - 1

Or, y = - 1 - 1

I.e y = -2

So, For x = 5 , y = - 2

Point is ( $x_1$ , $y_1$ ) = (5 , - 2)

Again , put the value of x in eq 2

So, y = $\dfrac{4}{5}$ x - 6

Or, y = $\dfrac{4}{5}$ × 5 - 6

Or, y = $\frac{4\times 5}{5}$ - 6

Or, y = 4 - 6

I.e y = - 2

So, For x = 5 , y = - 2

Point is ( $x_2$ , $y_2$ ) = (5 , - 2)

Hence, The points for the given to linear equations is (5 , - 2) and (5 , - 2)

The points is plotted on the graph shown . Answer

5 0

4 years ago

Find the surface area of a sphere with a volume of 36π in3.

swat32

First use known volume to solve for radius:
$V=\frac{4}{3}\pi r^{3}$
$r=\sqrt[3]{\frac{3V}{4\pi}}$
$r=\sqrt[3]{\frac{3(36\pi)}{4\pi}}$
$r=\sqrt[3]{27}$
$r=3$

Then use the radius to get the surface area:
$A_{surf}=4 \pi r^{2}$
$A_{surf}=4 \pi 3^{2}$
$A_{surf}=4 \pi 9$
$A_{surf}=36\pi\approx113.09$

6 0

3 years ago

What is mAngleMHJ? HELP pLEASE 35° 50° 72.5° 92.5°

telo118 [61]

Answer:

50°

Step-by-step explanation:

1) Opposite angles are the same. Therefore,

2x-20=x+15

2x-x=15+20

x=35

2) Put x back in the equation for angle mhj

2(35)-20=50

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3 years ago

F(t)=t^2+19t+60 What are the zeros of the function and what is the vertex of the parabola?

SVEN [57.7K]

Answer: $-4,-15;\ \left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$

Step-by-step explanation:

Given

Function is $F(t)=t^2+19+60$

Zeroes of the function are

$t^2+19t+60=0\\\\\Rightarrow t=\dfrac{-19\pm\sqrt{19^2-4\times 1\times 60}}{2\times 1}\\\\\Rightarrow t=\dfrac{-19\pm \sqrt{121}}{2}\\\\\Rightarrow t=\dfrac{-19\pm 11}{2}\\\\\Rightarrow t=-4,-15$

Using completing the square method

$y=t^2+2\times \dfrac{19}{2}t+\dfrac{19^2}{2^2}-\dfrac{19^2}{2^2}+60\\\\y=\left(t+\dfrac{19}{2}\right)^2+60-\dfrac{361}{4}\\\\y=\left(t+\dfrac{19}{2}\right)^2-\dfrac{121}{4}\\\\y+\dfrac{121}{4}=\left(t+\dfrac{19}{2}\right)^2\\\\\left(y-(-\dfrac{121}{4}\right)=\left(t-(-\dfrac{19}{2})\right)^2\\\\\text{The vertex is }\left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$

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3 years ago

How to find the steeper slope in a graph?​

How to find the steeper slope in a graph?