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2 years ago

Solve algebraically the simulaneous equations y=2x^2-3x-102x-y=-2

Mathematics

1 answer:

enot [183]2 years ago

3 0

Answer:

X=-3/2, X=4

Step-by-step explanation:

Really easy

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If f(x) = x^3 – 10x^2 + 29x – 30 and f(6) = 0, then find all<br> of the zeros of f(x) algebraically.

tino4ka555 [31]

Step 1: 1

(((x3) - (2•5x2)) + 29x) - 30 = 0

Step 2 2.1 x3-10x2+29x-30 is not a perfect cube

Step 3 Factoring: x3-10x2+29x-30

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1: 29x-30
Group 2: -10x2+x3

Pull out from each group separately :

Group 1: (29x-30) • (1)
Group 2: (x-10) • (x2)

7 0

2 years ago

What are the coordinates of the midpoint of the line segment with the endpoints (2,-5) and (8,3)

Papessa [141]

Midpoint Formula: $(\frac{x_2+x_1}{2} ,\frac{y_2+y_1}{2} )$

Plug the coordinates into the midpoint formula and solve as such:

$(\frac{2+8}{2} ,\frac{-5+3}{2} )\\\\(\frac{10}{2} ,\frac{-2}{2} )\\\\(5,-1)$

<u>Your midpoint is (5,-1).</u>

7 0

2 years ago

5. (9r 3 + 5r 2 + 11r) + (-2r 3 + 9r - 8r 2)

ad-work [718]

Answer:

7r³ - 3r² + 20r

Step-by-step explanation:

(9r³ + 5r² + 11r) + (-2r³ + 9r - 8r²)

You can get rid of the parentheses since there is nothing to distribute.

9r³ + 5r² + 11r - 2r³ + 9r - 8r²

Combine like factors.

7r³ - 3r² + 20r

3 0

2 years ago

Find the values of A B C AND D of <br> 4x^2 (2x^3+5x)= Ax^B +Cx^D

iren2701 [21]

The values are $A=8, B=5, C= 20$ and $D=3$

Explanation:

The expression is $4x^{2} (2x^{3} +5x)=Ax^{B} +Cx^{D}$

Simplifying, we get,

$8x^{5} +20x^3=Ax^{B} +Cx^{D}$

Since, both sides of the expression are equal, we can equate the corresponding values of A, B, C and D.

Thus, we get,

$8 x^{5}=A x^{B}$ ⇒ $A=8$ and $B=5$

Also, equating, $20 x^{3}=C x^{D}$ , we get,

$C=20$ and $D=3$

Thus, the values are $A=8, B=5, C= 20$ and $D=3$

3 0

2 years ago

Find the geometric sum: 40 + 20 + 10 + … + 0.078125

sveticcg [70]

$S=40+20+10+\cdots+0.078125$
$S=40\left(1+\dfrac12+\dfrac14+\cdots+\dfrac1{512}\right)$
$S=40\left(1+2^{-1}+2^{-2}+\cdots+2^{-9}\right)$

$2^{-1}S=40\left(2^{-1}+2^{-2}+2^{-3}+\cdots+2^{-10}\right)$

$\implies S-2^{-1}S=40\left(1-2^{-10}\right)$
$\implies\dfrac12S=40\left(1-\dfrac1{1024}\right)$
$\implies S=80\left(1-\dfrac1{1024}\right)=\dfrac{5115}{64}=79.921875$

4 0

2 years ago

Solve algebraically the simulaneous equations y=2x^2-3x-102x-y=-2​

Solve algebraically the simulaneous equations y=2x^2-3x-102x-y=-2