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1 year ago

A, B, C, D are points on a line.

Mathematics

1 answer:

mihalych1998 [28]1 year ago

6 0

The points A, B, C and D are collinear points

The length of the segment AD is 7 units

<h3>How to determine the length AD?</h3>

The given parameters are:

AB = 1

BC = 2

CD = 4

Because the points are on the same line, then the following is the possible equation

AD = AB + BC + CD

Substitute known values

AD = 1 + 2 + 4

Evaluate the sum

AD = 7

Hence, the length of the segment AD is 7 units

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a line passes through the point (-3,-5) and has a slope of 5 write an equation in point slope form for this line

Tanzania [10]

Answer:

$y=5x+10$

Step-by-step explanation:

1) Plug numbers into slope formula to find $b$ .

$y=mx+b$

$-5=5(-3)+b$

2) Simplify to find $b$ .

$-5=-15+b$

$10=b$

3) Check work.

$-5=-15+10$

This statement is true.

4) Write equation.

$y=5x+10$

4 0

1 year ago

Ac=r+d for a solve equation for the indicated variable

konstantin123 [22]

$\boxed{a=\frac{r+d}{c}}$

<h2>Explanation:</h2>

Here we have the following expression:

$ac=r+d$

And the question asks for solving the equation for the variable $a$ . So we need to divide both sides of the equation by $c$ :

$ac=r+d \\ \\ \frac{ac}{c}=\frac{r+d}{c} \\ \\ \\ But: \\ \\ \frac{c}{c}=1 \\ \\ \\ So: \\ \\ \boxed{a=\frac{r+d}{c}}$

<h2>Learn more:</h2>

Solving equations: brainly.com/question/10643312

#LearnWithBrainly

5 0

2 years ago

3x-5+2x=15+4x-5 how do I solve this step by step

bezimeni [28]

3x - 5 + 2x = 15+ 4x - 5
5x - 5 = 10 + 4x
5x - 5 + 5 = 10 +5 +4x
5x = 4X + 15
5x - 4x = 4x - 4x + 15
x = 15

6 0

3 years ago

Solve the system by substitution.<br> y = 3x + 28<br> y = -4x

Mazyrski [523]

Answer:

(-4, 16)

Step-by-step explanation:

3 0

2 years ago

If the sum of the zereos of the quadratic polynomial is 3x^2-(3k-2)x-(k-6) is equal to the product of the zereos, then find k?

lys-0071 [83]

Answer:

Step-by-step explanation:

So I'm going to use vieta's formula.

Let u and v the zeros of the given quadratic in ax^2+bx+c form.

By vieta's formula:

1) u+v=-b/a

2) uv=c/a

We are also given not by the formula but by this problem:

3) u+v=uv

If we plug 1) and 2) into 3) we get:

-b/a=c/a

Multiply both sides by a:

-b=c

Here we have:

a=3

b=-(3k-2)

c=-(k-6)

So we are solving

-b=c for k:

3k-2=-(k-6)

Distribute:

3k-2=-k+6

Add k on both sides:

4k-2=6

Add 2 on both side:

4k=8

Divide both sides by 4:

k=2

Let's check:

$3x^2-(3k-2)x-(k-6) \text{ with }k=2$ :

$3x^2-(3\cdot 2-2)x-(2-6)$

$3x^2-4x+4$

I'm going to solve $3x^2-4x+4=0$ for x using the quadratic formula:

$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\frac{4\pm \sqrt{(-4)^2-4(3)(4)}}{2(3)}$

$\frac{4\pm \sqrt{16-16(3)}}{6}$

$\frac{4\pm \sqrt{16}\sqrt{1-(3)}}{6}$

$\frac{4\pm 4\sqrt{-2}}{6}$

$\frac{2\pm 2\sqrt{-2}}{3}$

$\frac{2\pm 2i\sqrt{2}}{3}$

Let's see if uv=u+v holds.

$uv=\frac{2+2i\sqrt{2}}{3} \cdot \frac{2-2i\sqrt{2}}{3}$

Keep in mind you are multiplying conjugates:

$uv=\frac{1}{9}(4-4i^2(2))$

$uv=\frac{1}{9}(4+4(2))$

$uv=\frac{12}{9}=\frac{4}{3}$

Let's see what u+v is now:

$u+v=\frac{2+2i\sqrt{2}}{3}+\frac{2-2i\sqrt{2}}{3}$

$u+v=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}$

We have confirmed uv=u+v for k=2.

4 0

2 years ago