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3 years ago

Urgently needed see image

Mathematics

1 answer:

Lorico [155]3 years ago

7 0

y2-y1 =M(x2-x1)

Ok Sir, you gave points : (2,1) and (3,4)

4-1/3-2 = 3

Ok we know our slope is 3, now pick any of the two points, and make an equation for this line, so lets go ahead and pick #1, (2,1)

Formula is same as before

y-1=3x-6

y=3x-5

I think its correct, pick as brainless sir, thanks.

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What are the solutions to the quadratic equation below x^2+20x+100=7

aleksandrvk [35]

First you must have the quadratic equal to zero. In order to do this you must subtract 7 to both sides

x^2 + 20x + (100 - 7) = 7 - 7

x^2 + 20x + 93 = 0

Now you must find two numbers who's sum equals 20 and their multiplication equal 93

Are there any? NO!

This means that you have to use the formula:

$\frac{-b±\sqrt{b^{2} - 4ac} }{2a}$

In this case:

a = 1

b = 20

c = 93

$\frac{-(20) plus/minus\sqrt{20^{2} - 4(1)(93)} }{2*1}$

$\frac{-20 plus/minus\sqrt{400 - 372} }{2}$

$\frac{-20 plus/minus\sqrt{28} }{2}$

^^^We must simplify √28

√28 = 2√7

so...

$\frac{-20 plus/minus 2\sqrt{7} }{2}$

simplify further:

$-10 plus/minus\sqrt{7$

-10 + √7

-10 - √7

***plus/minus = ±

Hope this helped!

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3 years ago

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liberstina [14]

Answer:

the only solution is 0

Step-by-step explanation:

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3 years ago

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PLEASE HELP ME -45= -2.5g

Rainbow [258]

Answer:

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Step-by-step explanation:

Isolate the variables by dividing each side by -2.5.

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3 years ago

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Crickets can jump with a vertical velocity of up to 14 ft/s. Which equation models the height of such a jump, in feet, after t s

Y_Kistochka [10]

Answer:

h = 147 - 16t^2.

Step-by-step explanation:

Use the following equation of motion

h = ut + 1/2 gt^2 where h = height, u = initial velocity , g = acceleration due to gravity and t = the time.

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h = 14t + 1/2 * -32 * t^2

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3 years ago

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Let f (x) = 3x − 1 and ε > 0. Find a δ > 0 such that 0 < ∣x − 5∣ < δ implies ∣f (x) − 14∣ < ε. (Find the largest

s344n2d4d5 [400]

Answer:

This proves that f is continous at x=5.

Step-by-step explanation:

Taking f(x) = 3x-1 and $\varepsilon>0$ , we want to find a $\delta$ such that $|f(x)-14|$

At first, we will assume that this delta exists and we will try to figure out its value.

Suppose that $|x-5|$ . Then

$|f(x)-14| = |3x-1-14| = |3x-15|=|3(x-5)| = 3|x-5|< 3\delta$ .

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By definition, this procedure proves that $\lim_{x\to 5}f(x) = 14$ . Note that f(5)=14, so this proves that f is continous at x=5.

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3 years ago