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Katyanochek1 [597]
3 years ago
12

What is the 9th term of -5,10,-20

Mathematics
1 answer:
slega [8]3 years ago
7 0
The answer is -1280
hope this helps
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Calculate the derivative of the following function. y=e6x(6x+1)5
netineya [11]

Answer:

Y' = 360ex+ 30e

Step-by-step explanation:

Y=e×6x×(6x+1) × 5

y'= d/dx (e×6x ×(6x+1)×5)

y'=d/dx(30ex ×(6x+1))

y'=d/dx(180ex^2 + 30ex)

y'=d/dx(180ex^2) + d/dx (30ex)

Calculate the derivative

y'=180e×2x +30e

y'=360ex+30e

6 0
3 years ago
What is the leading coefficient in the quadratic below?<br> h(x)=-2x2+5x+10
Mars2501 [29]

Answer:

-2

Step-by-step explanation:

The leading coefficient in a polynomial is the coefficient that corresponds with the highest degree term. In this case, -2 is the leading coefficient since its term has x² as the highest degree of the quadratic.

5 0
2 years ago
What is the answer ​
aliya0001 [1]

Answer:

(4,5)

(6,2)

(2,4)

Step-by-step explanation:

just add 3 to x and 1 to y

3 0
3 years ago
Read 2 more answers
PLS HELP ME ON THIS QUESTION I WILL MARK YOU AS BRAINLIEST IF YOU KNOW THE ANSWER PLS GIVE ME A STEP BY STEP EXPLANATION!!
uysha [10]

Answer:

B) 9

Step-by-step explanation:

Quotient means that it’s for DIVISION. The 18 is the dividend, 2 is the divisor, and the quotient is the answer:

Dividend ÷ Divisor = Quotient

18 ÷ 2 = 9

3 0
3 years ago
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
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