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padilas [110]
3 years ago
15

A line that includes the point ( - 4, - 1) has a slope of - 2. What is its equation in point-slope form?

Mathematics
2 answers:
EastWind [94]3 years ago
4 0

Point-Slope equation: y - y₁ = m(x - x₁)  ; m is slope and (x₁ ,y₁) is the coordinate

y - (-1) = -2(x - (-4)

y + 1 = -2(x + 4)     <em>simplified (two negatives make a positive)</em>


Grace [21]3 years ago
3 0

Point slope form says the line through (a,b) with slope m is

y - b = m(x - a)

Here that's

y - - 1 = -2(x - -4)

Answer: y + 1 = -2(x + 4)



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2 years ago
Write a quadratic function in standard form with zeros 1 and -10
Licemer1 [7]

Answer:

f(x)=x^2+9x-10

Step-by-step explanation:

<u>Standard Form of Quadratic Function</u>

The standard form of a quadratic function is:

f(x)=ax^2+bx+c

where a,b, and c are constants.

The factored form of a quadratic equation is:

f(x)=a(x-\alpha)(x-\beta)

Where \alpha and \beta are the roots or zeros of f, and a is constant.

We know the zeros of the function are 1 and -10. The function is:

f(x)=a(x-1)(x-(-10))

f(x)=a(x-1)(x+10)

Operating:

f(x)=a(x^2+10x-x-10)

Joining like terms:

f(x)=a(x^2+9x-10)

Since we are not given any more restrictions, we can choose the value of a=1, thus. the required function is:

\boxed{f(x)=x^2+9x-10}

6 0
3 years ago
In your own words, define Quadratic Equation. How many solutions does a Quadratic Equation have?
dmitriy555 [2]

Answer: an equation that has one term which is nameless and squared also no term which gets raised to higher power.

Step-by-step explanation:

5 0
2 years ago
In a poll of Mr. Briggs's math class, 67% of the students say that math is their favorite academic subject. The editor of the sc
levacccp [35]

Mathematics is the subject that is most popular in the class and is liked by the students.

<u>Explanation:</u>

Among all the subjects that are added in the curriculum of the students in the school, Mathematics is the subject that is mostly liked by the students in the class and in the school.

To find this that mathematics was the favorite subject of the students, a a survey was conducted by the school people in the class of a professor and through that survey it was shown that Mathematics was a subject that was mostly liked by all the students.

4 0
3 years ago
Please help me thank you
Hoochie [10]

Answer:

\large\boxed{\sin2\theta=\dfrac{\sqrt3}{2},\ \cos2\theta=\dfrac{1}{2}}

Step-by-step explanation:

We know:

\sin2\theta=2\sin\theta\cos\theta\\\\\cos2\theta=\cos^2\theta-\sin^2\thet

We have

\sin\theta=\dfrac{1}{2}

Use \sin^2\theta+\cos^2\theta=1

\left(\dfrac{1}{2}\right)^2+\cos^2\theta=1\\\\\dfrac{1}{4}+\cos^2\theta=1\qquad\text{subtract}\ \dfrac{1}{4}\ \text{from both sides}\\\\\cos^2\theta=\dfrac{4}{4}-\dfrac{1}{4}\\\\\cos^2\theta=\dfrac{3}{4}\to\cos\theta=\pm\sqrt{\dfrac{3}{4}}\to\cos\theta=\pm\dfrac{\sqrt3}{\sqrt4}\to\cos\theta=\pm\dfrac{\sqrt3}{2}\\\\\theta\in[0^o,\ 90^o],\ \text{therefore all functions have positive values or equal 0.}\\\\\cos\theta=\dfrac{\sqrt3}{2}

\sin2\theta=2\left(\dfrac{1}{2}\right)\left(\dfrac{\sqrt3}{2}\right)=\dfrac{\sqrt3}{2}\\\\\cos2\theta=\left(\dfrac{\sqrt3}{2}\right)^2-\left(\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{3-1}{4}=\dfrac{2}{4}=\dfrac{1}{2}

3 0
2 years ago
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