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3 years ago

You rent an apartment that costs $1800 per month during the first year, but the rent

Mathematics

1 answer:

RUDIKE [14]3 years ago

5 0

Answer:

2710 is the answer

Step-by-step explanation:

1800+130(7)

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PLEASE HELP

Ede4ka [16]

I believe the answer is B. 29x + y = 348

6 0

4 years ago

NO LINKS PLEASE HURRY I REALLY NEED HELP

alisha [4.7K]

Answer:

The choice;

$- 1.45x - 3.41$

I hope I helped you^_^

4 0

3 years ago

Read 2 more answers

If α and β are the zeroes of the polynomial 6y 2 − 7y + 2, find a quadratic polynomial whose zeroes are 1 α and 1 β .

ollegr [7]

Answer:

$2y^2-7y+6=0$

Step-by-step explanation:

We are given that $\alpha$ and $\beta$ are the zeroes of the polynomial $6y^2-7y+2$

$y^2-\frac{7}{6}y+\frac{1}{3}$

We have to find a quadratic polynomial whose zeroes are $1/\alpha$ and $1/\beta$ .

General quadratic equation

$x^2-(sum\;of\;zeroes)x+ product\;of\;zeroes$

We get

$\alpha+\beta=\frac{7}{6}$

$\alpha \beta=\frac{1}{3}$

$\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}$

$\frac{1}{\alpha}+\frac{1}{\beta}=\frac{7/6}{1/3}$

$\frac{1}{\alpha}+\frac{1}{\beta}=\frac{7}{6}\times 3=7/2$

$\frac{1}{\alpha}\times \frac{1}{\beta}=\frac{1}{\alpha \beta}$

$\frac{1}{\alpha}\times \frac{1}{\beta}=\frac{1}{1/3}=3$

Substitute the values

$y^2-(7/2)y+3=0$

$2y^2-7y+6=0$

Hence, the quadratic polynomial whose zeroes are $1/\alpha$ and $1/\beta$ is given by

$2y^2-7y+6=0$

4 0

3 years ago

What is 32/5 times 11/4?

Veronika [31]

The answer is 88/5, which is 17.6.

8 0

3 years ago

Read 2 more answers

In an algebra 2 test and I'm stumped

quester [9]

7/3-<span>1.41
7/1.59
4.4
That's what i got...

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3 0

4 years ago

Read 2 more answers