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2 years ago

What is the slope-intercept equation of the line below? 61 -5

Mathematics

1 answer:

Lemur [1.5K]2 years ago

7 0

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

The Correct option is ~ B

$\boxed{ \boxed{y = 2x - 3}}$

$\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}$

Let's find the slope (m) using points (2 , 1) and (0 , -3)

$\mathrm{ \dfrac{y2 - y1}{x2 - x1} }$

$\dfrac{1 - ( - 3)}{2 - 0}$

$\dfrac{4}{2}$

hence, slope = 2

now, by Observing the given graph we can infer that the given line cuts the y - axis at point (0 , -3), so value of y - intercept (c) = - 3

[ y - coordinate of a point when x - coordinate is equal to 0 is the value of y - intercept of a line ]

And, we know the general equation of line in slope - intercept form is ~

$\boxed{y = mx + c}$

now, let's plug the value of slope (m) and y - intercept (c) in the general equation to find the equation of line in slope intercept form ~

$y = 2x - 3$

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What are all of the real roots of the following polynomial? f(x) = x5 + 5x4 - 5x3 - 25x2 + 4x + 20

Nadusha1986 [10]

F(x)=x^5 + 5*x^4 - 5*x^3 - 25*x^2 + 4*x + 20

By examining the coefficients of the polynomial, we find that
1+5-5-25+4+20=0 => (x-1) is a factor
Now, reverse the sign of coefficients of odd powers,
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By the rational roots theorem, we can continue to try x=2, or factor x-2=0
2^5+5(2^4)-5(2^3)-25(2^2)+4(2)+20=0
and similarly f(-2)=0
So we have found four of the 5 real roots.
The remainder can be found by synthetic division as x=-5

Answer: The real roots of the given polynomial are: {-5,-2,-1.1.2}

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3 years ago

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Talja [164]

Answer:

the number of possibilities that can be drace himself up for the day is 216

Step-by-step explanation:

Given that

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So here

$= 12C_1 \times 6C_1 \times 3C_1\\\\= \frac{12!}{1!(12 - 1)!} \times \frac{6!}{1!(6 - 1)!} \times \frac{3!}{1!(3 - 1)!}\\\\= \frac{12!}{1!(11)!} \times \frac{6!}{1!(5)!} \times \frac{3!}{1!(2)!}\\\\= 12 \times 6 \times 3\\\\= 216$

Hence, the number of possibilities that can be drace himself up for the day is 216

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2 years ago

Is it possible to have a triangle such that none of the exterior angles is obtuse? Please help and explain

PolarNik [594]

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The other two interior angles must be acute if you go with 1 obtuse angle, which leaves the corresponding exterior angles to be obtuse. If you wanted the exterior angles to be all acute, then you would need 3 obtuse interior angles, but that is not possible.

Why isn't it possible to have 3 obtuse angles? Consider a triangle with interior angles A, B, C. The three angles must add to 180

A+B+C = 180

If A,B,C were all larger than 90, then

A > 90

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A+B+C > 90+90+90

A+B+C > 270

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3 years ago

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lisov135 [29]

Answer:

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3 years ago

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ExtremeBDS [4]

Answer:

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$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

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e = 2.71828 is the Euler number

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In this question:

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2 years ago