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3 years ago

6x^{2} + 19 + 9x^{2} - 8 With Explanation

Mathematics

1 answer:

NemiM [27]3 years ago

7 0

Answer:

I'mnot sure if you're asking for its roots..but here you go :)

Step-by-step explanation:

$= 6 {x}^{2} + 19 + 9 {x}^{2} - 8 \\ = (6 {x}^{2} + 9 {x}^{2} ) + (19 - 8) \\ = 15 {x}^{2} + 8 \\ 15 {x}^{2} = - 8 \\ {x}^{2} = - \frac{8}{15} \\ \sqrt{ {x}^{2} } = \sqrt{ - \frac{8}{15} } \\ x = \sqrt{ \frac{8}{15}} \times \sqrt{ - 1} \\ x = \sqrt{ \frac{8}{15} } \times i \\ x1 = 0.7302i \\ x2 = - 0.7302i$

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If f(x) = 3x-2, then f (8) - f (-5) =

Volgvan

Step-by-step explanation:

= f (8) - f (-5)

= 3*8-2 -(3*-5-2)

= 22-(-15-2)

= 22+17

= 39

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4 years ago

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What is the first term of the polynomial Raj ended up with? 6x3 8x3 2xy2 3xy2

laila [671]

The polynomial is a third degree polynomial with x and y terms so when written in standard form x comes before y and the x^3 is going to be the first term. The first term of this polynomial is 8x^3, aka the second answer choice.

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3 years ago

Solve for x: 4 over x plus 4 over quantity x squared minus 9 equals 3 over quantity x minus 3. (2 points) Select one: a. x = -4

Kay [80]

Answer:

Step-by-step explanation:

$\dfrac{4}{x}+\dfrac{4}{x^2-9}=\dfrac{3}{x-3}$

Domain:

$x\neq0\ \wedge\ x^2-9\neq0\ \wedge\ x-3\neq0\\\\x\neq0\ \wedge\ x\neq\pm3$

solution:

$\dfrac{4}{x}+\dfrac{4}{x^2-3^2}=\dfrac{3}{x-3}$

use (a - b)(a + b) = a² - b²

$\dfrac{4}{x}+\dfrac{4}{(x-3)(x+3)}=\dfrac{3}{x-3}$

multiply both sides by (x - 3) ≠ 0

$\dfrac{4(x-3)}{x}+\dfrac{4(x-3)}{(x-3)(x+3)}=\dfrac{3(x-3)}{x-3}$

cancel (x - 3)

$\dfrac{4(x-3)}{x}+\dfrac{4}{x+3}=3$

subtract $\frac{4(x-3)}{x}$ from both sides

$\dfrac{4}{x+3}=3-\dfrac{4(x-3)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x}{x}-\dfrac{(4)(x)+(4)(-3)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x-\bigg(4x-12\bigg)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x-4x-(-12)}{x}\\\\\dfrac{4}{x+3}=\dfrac{-x+12}{x}$

cross multiply

$(4)(x)=(x+3)(-x+12)$

use FOIL

$4x=(x)(-x)+(x)(12)+(3)(-x)+(3)(12)\\\\4x=-x^2+12x-3x+36$

subtract 4x from both sides

$0=-x^2+12x-3x+36-4x$

combine like terms

$0=-x^2+(12x-3x-4x)+36\\\\0=-x^2+5x+36$

change the signs

$x^2-5x-36=0\\\\x^2-9x+4x-36=0\\\\x(x-9)+4(x-9)=0\\\\(x-9)(x+4)=0$

The product is 0 if one of the factors is 0. Therefore:

$x-9=0\ \vee\ x+4=0$

$x-9=0$ add 9 to both sides

$x=9\in D$

$x+4=0$ subtract 4 from both sides

$x=-4\in D$

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3 years ago

What two integers are between √88

MariettaO [177]

9 and 10
hope this helps!!!

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3 years ago

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HEL ME PLEEAASSEE AMIGO PLEEAASSEE or amiga

podryga [215]

Answer:

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Step-by-step explanation:

A full semicircle (or half circle) has a degrees of 180. And x is equal that semicircle so x = 180, now if we subtract 21 from 180 we get 159, so X = 159.

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3 years ago