Can u choose if it is primary or secondary please and get it right.

Can you guys help me!!!! Question 3

Makovka662 [10]

Answer:

F = ( 9/5) X c + 32

Step-by-step explanation:

5 0

3 years ago

PLEASE HELP!!!<br> Calculate the length of segment with the midpoint.

andrezito [222]

Answer:

17. 10

Step-by-step explanation:

1. A segment going from an endpoint to the midpoint of the original segment is going to be 1/2 of the original segment.

AM = 1/2 AB

2. You know that the length of AM is 5, so plug that in a solve algebraically

5 = 1/2 AB

(2)5 = (2) 1/2 AB

10 = AB

Answer:

18. 30

Step-by-step explanation:

The sum of two segments spanning from the original segment's midpoint to the end equals the length of the original segment. Because the midpoint is exactly in the middle of the original segment, the two other segments should equal each other.

1. You need to first find the length of the two segments by setting them equal to each other and plugging in their equations.

5x = x+12

2. Solve algebraically

5x = x+12

4x = 12

x = 3

3. Plug z into the equations for each segment and add them together.

RM = 5(3) MS = (3)+12

RM = 15 MS = 15

15+15 = 30

4 0

3 years ago

Read 2 more answers

Easy question

Anarel [89]

Answer:

d)33

e)80

Step-by-step explanation:

I'm sorry I don't really know how to explain this.

5 0

3 years ago

Read 2 more answers

Jaz was 43 inches tall. Eighteen months later, she was 52 inches tall. The constant rate of change for Jaz's height was .

maxonik [38]

Answer:

Constant rate of change for Jaz height will be 0.5 inch per month

Step-by-step explanation:

Jaz present height is 43 inches.

Eighteen months later her height will be 52 inches.

Now we know rate of change will be represented by the formula

Rate of change of height =

=

= 0.5 inch per month

The constant rate of change for Jaz height will be 0.5 inch per month.

Read more on Brainly.com - brainly.com/question/2012372#readmore

7 0

3 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:

2

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

3 years ago