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11111nata11111 [884]
3 years ago
10

Find the value of x2 for these values of x a: 2 b: 3 c:9 d:10 Help Please?

Mathematics
1 answer:
larisa86 [58]3 years ago
4 0
X=1 so you need to add and it will be c=9
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Which of the following is a solution of x2 + 6x = −18? x = 3 − 3i x = −3 + 3i x = −6 + 3i x = 6 − 3i
Marina86 [1]

Step 1 ) Move all terms right side of the equation.

\displaystyle\ x^{2} + 6x = -18

\displaystyle\ x^{2} + 6x + 18 = -18 + 18

\displaystyle\ x^{2} +6x + 18 = 0

Step 2 ) Apply quadratic formula. (Note: There are 2 solutions)

\displaystyle\ x^{2} +6x + 18 = 0

\displaystyle\ x = \frac{-b\sqrt{b^{2}-4ac}}{2a}, \displaystyle\ x = \frac{-b-\sqrt{b^{2}-4ac}}{2a}

\displaystyle\ x = \frac{-6+\sqrt{6^{2}-4 * 18}}{2} , \displaystyle\ x = \frac{-6-\sqrt{6^{2}-4*18}}{2}

\displaystyle\ x = \frac{-6+6i}{2} ,

\displaystyle\ x = \frac{-6-6i}{2}

Step 3 ) Simplify.

\displaystyle\ x = \frac{-6+6i}{2} ,

\displaystyle\ x = \frac{-6-6i}{2}

\displaystyle\ x = -3+ 3i,

\displaystyle\ x = -3 - 3i

Since the options only provide one of the answer we found, the answer is...

\displaystyle\ x = -3 - 3i

•

•

- <em>Marlon Nunez</em>

6 0
3 years ago
Read 2 more answers
I need some help with this. If you could help me, that would be great.
Vesnalui [34]

Answer:

 .31 g / mL

Step-by-step explanation:

Density = mass / volume

Sample 3

mass = 7 1/4 = 7.25

Volume 23.1

Density = 7.25/ 23.1 = .313852814

To the nearest hundredth

  .31 g / mL

8 0
3 years ago
PLZ HURRY IT'S URGENT!!!!!!
Anton [14]

I think the correct answer your looking for is D) 9

5 0
3 years ago
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For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4
Elina [12.6K]

Answer:

<em>C.</em> (5-\frac{1}{2})^6

Step-by-step explanation:

Given

15(5)^2(-\frac{1}{2})^4

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

Sum = 2 + 4

Sum = 6

Each term of a binomial expansion are always of the form:

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

Where n = the sum above

n = 6

Compare 15(5)^2(-\frac{1}{2})^4 to the above general form of binomial expansion

(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......

Substitute 6 for n

(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......

[Next is to solve for a and b]

<em>From the above expression, the power of (5) is 2</em>

<em>Express 2 as 6 - 4</em>

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

By direct comparison of

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

and

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

We have;

^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4

Further comparison gives

^nC_r = 15

a^{n-r} =(5)^{6-4}

b^r= (-\frac{1}{2})^4

[Solving for a]

By direct comparison of a^{n-r} =(5)^{6-4}

a = 5

n = 6

r = 4

[Solving for b]

By direct comparison of b^r= (-\frac{1}{2})^4

r = 4

b = \frac{-1}{2}

Substitute values for a, b, n and r in

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

Solve for ^6C_4

(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......

<em>Check the list of options for the expression on the left hand side</em>

<em>The correct answer is </em>(5-\frac{1}{2})^6<em />

3 0
3 years ago
Charges competition fees of $250 plus $75 per week for classes. What is the
Nimfa-mama [501]

Answer:

$3,250

Step-by-step explanation:

You take the initial fee of 250 then add 75x40

250x75(40)=$3250

3 0
2 years ago
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