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3 years ago

Can someone please help me with this one? Tysm!!

Mathematics

2 answers:

kirill115 [55]3 years ago

4 0

Answer:

-26

Step-by-step explanation:

well add the negative numbers first so you get

-17-18=(-35)

then you add 9 so its (-35)+9=26

Marina86 [1]3 years ago

3 0

Answer:

Step-by-step explanation:

-17 + 9 +(-18) = -8 -18 = -26 final answer

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Solve. | 3x + 1 | = 2

qwelly [4]

Since this is an absolute value equation, it will have two answers. For the first answer, take away the absolute value bars and solve 3x + 1 = 2. Subtract 1 from both sides to get 3x = 1 and divide each side by 3 to get x = 1/3. Now onto the second solution. This time, take away the absolute value bars and make the other side of the equation, the 2, negative, to get 3x + 1 = -2. Now solve this by subtracting 1 from each side to get 3x = -3 and divide each side to get the other answer which is x = -1. The answer is x = -1 or 1/3, hope this helps!

8 0

3 years ago

Read 2 more answers

Monica knows that she should spend no more than 1/3 of her income on her rent. She really wants to move into a cool condo downto

Andrew [12]

Answer:

$6,900

Step-by-step explanation:

Let

x = Amount Monica should earn each month

Rent per month = $2,300

Monica knows that she should spend no more than 1/3 of her income on her rent

1/3 of her Income = rent

1/3 of x = 2,300

1/3x = 2,300

x = 2,300 ÷ 1/3

x = 2,300 * 3/1

x = $6,900

Monica should earn $6,900 each month to afford the rent on her cool new place

7 0

3 years ago

Can someone please help on my geometry test?

sveta [45]

THE ANSWER IS NOT IN THE LINK THAY THAT PERSON GAVE

5 0

3 years ago

4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.

nikitadnepr [17]

Answer:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=1$

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0$

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}$

Step-by-step explanation:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))$

$=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})$

$=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1$

2nd

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}$

$=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0$

3th

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))$

$=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}$

What we use?

We use that

$e^{i\pi n}=cos(\pi n)+i sin(\pi n)$

and

$\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}$

6 0

3 years ago

PLEASE HELP ME!! BRAIINLIEST WILL BE GIVEN

PtichkaEL [24]

Your answer would be answer choice C. The initial number of transactions is a dependent variable, because the number of transactions made are dependent on the number of hours that have passed.

Hope this helps,
♥A.W.E.S.W.A.N.♥

4 0

3 years ago