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3 years ago

Please help me ASAPP

Mathematics

1 answer:

swat323 years ago

7 0

Answer:

I believe the answer is, A.

Explanation:

Ordered pairs are classified as a function when none of the x’s in the ordered pairs match.

Like in b, a pair has 3.
And in c, two pairs have 1 and 3.

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Please help :(

faltersainse [42]

I think the answer is c

7 0

3 years ago

Lisa pays $1339.70 quarterly on her property taxes, what is her monthly payment?

Serga [27]

Every quarter is 3 months so you must divide 1339.70 by 3 to get her monthly payment of about $446.57.

Hope this helps :3

7 0

3 years ago

Solve for c. -3b+ac=c-4. Please show all work.

Luden [163]

-3b+ac=4
Move c to the left and - 3b to the right
ac-c=-4+3b
Now factor by c at the left
c(a-1)=-4+3b
Divide the right by a-1
C=(-4+3b)/(a-1)

6 0

3 years ago

Read 2 more answers

Answer question 1-15 by 6pm

alina1380 [7]

Answer:

Step-by-step explanation:

It is not really a question we can help you with you have to draw shapes for each 3 letters

8 0

3 years ago

Q(a+1) - 2q(a) if q(x) = x²+3x+4

siniylev [52]

$q(x) = {x}^{2} + 3x + 4$

_________________________________

Step(1)

To find q(a) we just need to put a instead of x in q(x) function.

Let's do it...

$q(a) = {a}^{2} + 3a + 4$

Multiply sides by -2 :

$- 2q(a) = - 2( {a}^{2} + 3a + 4)$

$- 2q(a) = - 2 {a}^{2} - 6a - 8$

_________________________________

Step (2)

To find q(a+1) we just need to put a+1 instead of x in q(x) function.

Let's do it...

$q(a + 1) = ({a + 1})^{2} + 3(a + 1) + 4 \\$

$q(a + 1) = {a}^{2} + 2a + 1 + 3a + 3 + 4 \\$

$q(a + 1) = {a}^{2} + 5a + 8$

_________________________________

Step (3)

$q(a + 1) - 2q(a) =$

${a}^{2} + 5a + 8 - 2 {a}^{2} - 6a - 8 = \\$

$- {a}^{2} - a = - a(a + 1)$

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

8 0

4 years ago