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2 years ago

Two pages I need help on :)

Mathematics

1 answer:

aleksandr82 [10.1K]2 years ago

3 0

Answer:

Sorry for the wait...here's the solutions!

Download pdf

You might be interested in

If a number a is chosen at random from the set

Ivanshal [37]

Answer:

D. 3/19

Step-by-step explanation:

12/1 = 12

12/2 = 6

12/3 = 4

These are the only ones that make the equation true.

There are also only 19 numbers because they need to be less than 20 like the question indicated. Therefore, the answer is D.

5 0

3 years ago

Please help me with these two problems thank you!!

andriy [413]

Answer:

1st question: y=1/3x-1

2nd question: y=-2/3x+9

Step-by-step explanation:

Y=mx+b form is what you are using

when a line in parallel to a line they have the same slop wich is m

when a line in perpendicular they have the opposite reciprocal

Examples: 2/5 opposite reciprocal would be -5/2

8 0

3 years ago

I’m the parallelogram above, what is the value of x+y <br><br> Help please

olga2289 [7]

Answer:

248°

Step-by-step explanation:

y = 180°- 56°

= 124°

x = 124° because opposite angles of a parallelogram are equal

x + y = 124° + 124°

= 248°

6 0

3 years ago

Read 2 more answers

camie and jorge each saving money for a new computer camie saves $5 per week plus $50 she received for her birthday. jorge saves

ki77a [65]

Answer:

7w<u> > </u>5w + 50

Step-by-step explanation:

8 0

3 years ago

Find the equation, (f(x) = a(x - h)2 + k), for a parabola containing point (2, -1) and having (4, -3) as a vertex. What is the s

Nataliya [291]

Answer:

$f(x)=\frac{1}{2}x^2-4x+5$

Step-by-step explanation:

A parabola is written in the form

$f(x)=a((x-h)^2+k)$ (1)

where:

$h$ is the x-coordinate of the vertex of the parabola

$ak$ is the y-coordinate of the vertex of the parabola

$a$ is a scale factor

For the parabola in the problem, we know that the vertex has coordinates (4,-3), so we have:

$h=4$ (2)

$ak=-3$

From this last equation, we get that $a=\frac{-3}{k}$ (3)

Substituting (2) and (3) into (1) we get the new expression:

$f(x)=-\frac{3}{k}((x-4)^2+k) = -\frac{3}{k}(x-4)^2 -3$ (4)

We also know that the parabola contains the point (2,-1), so we can substitute

x = 2

f(x) = -1

Into eq.(4) and find the value of k:

$-1=-\frac{3}{k}(2-4)^2-3\\-1=-\frac{3}{k}\cdot 4 -3\\2=-\frac{12}{k}\\k=-\frac{12}{2}=-6$

So we also get:

$a=-\frac{3}{k}=-\frac{3}{-6}=\frac{1}{2}$

So the equation of the parabola is:

$f(x)=\frac{1}{2}((x-4)^2 -6)$ (5)

Now we want to rewrite it in the standard form, i.e. in the form

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

To do that, we simply rewrite (5) expliciting the various terms, we find:

$f(x)=\frac{1}{2}((x^2-8x+16)-6)=\frac{1}{2}(x^2-8x+10)=\frac{1}{2}x^2-4x+5$

6 0

3 years ago