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inysia [295]
3 years ago
10

Simplify by combining like terms 5 4/7i+4 3/5i

Mathematics
1 answer:
Arturiano [62]3 years ago
7 0
Simplify by combining the real and imaginary parts of each expression

Answer: 356i/35
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Need help with these math questions
PSYCHO15rus [73]

\sqrt{16-x}        x = 8

Since x = 8, you can plug in/substitute 8 for "x" in the equation:

\sqrt{16-x}

\sqrt{16-8}

\sqrt{8}

2.82842

2.83 You answer is the 4th option


\sqrt{x+7}     x = 9

Plug in 9 for "x" in the equation

\sqrt{x+7}

\sqrt{9+7}

\sqrt{16}

4    Your answer is the 1st option

4 0
3 years ago
Please help me I really need help
Oliga [24]

Answer:

Step-by-step explanation:

(1) 2x - 6y = -12

(2) x + 2y = 14

There is a -6y and a +2y. Since they have opposite signs, I'll try to eliminate the y terms. (That's my choice. There is more than one way to solve these.)

Multiply eq. (2) by 3:

3x + 6y = 42

Then add the result to eq. (1) to eliminate the y terms:

2x - 6y = -12

3x + 6y = 42

------------------

5x   = 30,  so x = 6

Now plug the value of x into eq. (2) and solve for y:

6 + 2y = 14

2y = 8

y = 4

Why did I use eq. (2) to solve for y? Because it's less work. I could have used eq. (1) instead:

2(6) - 6y = -12

12 - 6y = -12

-6y = -24

y = 4

More than one way to solve.

5 0
3 years ago
I will mark brain Can someone help me do not answer if u dont know
meriva

Answer:

1. y = 1/3x+-1

2. y = 4

Hope this helps man :D

6 0
3 years ago
Read 2 more answers
Find a number if 0.7 of it is 3.43.
lukranit [14]
Answer : I’m thinking 2.73 .

It shouldn’t be a trick question just fake 3.43 minus .7!
3 0
3 years ago
Write these series with summation notation. 1,4,9,16...
maksim [4K]

Answer:  \sum\limits_{i=1}^{n} n^2 , where n is a natural number.

Step-by-step explanation:

A series can be represented in a summation or sigma notation.

Greek capital letter, ∑ (sigma), is used to represent the sum.

For example: \sum\limits_{n=1}^{\infty} n=1+2+3+4+5+..., where n is a natural number.

The given series : 1,4,9,16 which can be written as 1^2, 2^2, 3^2,... .

So , we can write it as

\sum\limits_{n=1}^{\infty} n^2 , where n is a natural number.

7 0
3 years ago
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