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egoroff_w [7]
3 years ago
13

If x-3 and x-1/3 are both factors of of ax^2+5x+ab,then show that a=-3/2

Mathematics
1 answer:
faltersainse [42]3 years ago
8 0
Since x-3 and x-1/3 are factors

(x-3)(x-1/3)= x²-3x-(1/3)x+1= x²-(10/3)x+1

to make -(10/3)x to 5x multiply by -(3/10)*5= -3/2

-(3/2)(x-3)(x-1/3)= -(3/2)x²+5x-3/2

Now by comparing -(3/2)x²+5x-3/2 with <span>ax^2+5x+ab
</span>
you can see that a= -3/2
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Find the sum of the series.
ch4aika [34]

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as per my point to view you can use a L.C.M so you can get the answer or you can try using a exponent way like 5by 1 as n value is 1 so it may give yoh answer

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Write a quadratic function in standard form having zeros of -9 and -12
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(x + 9)(x + 12) = 0
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x^{2}+21x+108=0
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3 years ago
If the radius of the base is 2, and the height is 4, what is the volume of the cylinder?
MAVERICK [17]
givens
pi = 3.14
r = 2
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3 years ago
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Find the sum of a finite geometric sequence from n = 1 to n = 7, using the expression −4(6)n − 1.
Verizon [17]

Answer:

<h2>-223,948</h2>

Step-by-step explanation:

The formula of a sum of terms of a gometric sequence:

S_n=a_1\cdot\dfrac{1-r^n}{1-r}

a₁ - first term

r - common ratio

We have

a_n=-4(6)^{n-1}

Calculate a₁. Put n = 1:

a_1=-4(6)^{1-1}=-4(6)^0=-4(1)=-4

Calculate the common ratio:

r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=-4(6)^{n+1-1}=-4(6)^n\\\\r=\dfrac{-4(6)^n}{-4(6)^{n-1}}=6^n:6^{n-1}\\\\\text{use}\ a^n:a^m=a^{n-m}\\\\r=6^{n-(n-1)}=6^{n-n+1}=6^1=6

\text{Substitute}\ a_1=-4,\ n=7,\ r=6:\\\\S_7=-4\cdot\dfrac{1-6^7}{1-6}=-4\cdot\dfrac{1-279936}{-5}=-4\cdot\dfrac{-279935}{-5}=(-4)(55987)\\\\S_7=-223948

7 0
3 years ago
I need help please help me :(
zalisa [80]
EXPLANATION:
10 - ( 3 + 2*2 ) + 4
=> 10 - ( 3 + 4 ) + 4
=> 10 - 7 + 4
=> 3 + 4
=> 7

ANSWER: A. 7

Hope it helps u!
6 0
3 years ago
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