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2 years ago

3x - 9 + 4x Please Help

Mathematics

2 answers:

Wewaii [24]2 years ago

7 0

Answer:

Add like terms 7x - 9

Step-by-step explanation:

Sveta_85 [38]2 years ago

7 0

Add the 4 and 3 then you will get

7x-9

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A student writes the inequality x > 17 in interval notation as [17, POS INF]. Explain why this is incorrect.

krok68 [10]

Answer:

see explanation

Step-by-step explanation:

[ 17, + ∞ ] are used to indicate that the values inside are equal to the 2 limits, that is that x can equal 17 or + ∞

However, this as not the case as x > 17 indicates x is greater than 17 but not equal to it. This is also true for + ∞

To represent this situation parenthesis is used, that is

(17, + ∞ )

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3 years ago

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andreyandreev [35.5K]

Answer: 3

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3 years ago

When the polynomial 4x^4 − 13x^2 − 2x is divided by the polynomial x − 2 , x-2, what is the remainder?

zhenek [66]

Let p (x) = 4x^4-13x^2-2x
The zero of x-2 is 2
putting x = 2 in p (x), we get,
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3 years ago

if 1/5 and -2 are respectively product and sum of the zeroes of a quadratic polynomial. Find the polynomial.

Nikitich [7]

Answer:

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Step-by-step explanation:

To find: Quadratic Polynomial.

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3 years ago

Read 2 more answers

In two or three sentences explain how you would solve for the real solutions of the following equation: please help asap!!

g100num [7]

the real solutions for the equation $x^{3}=-64$ are -

$x=4,2+-2\sqrt{3i}$

Step-by-step explanation:

$x^{3}$ = $- 64$

$x^{3} +64$ = 0

We can write 64 as $4^{3}$

$x^{3}$ + $4^{3}$ = 0

using the identity ( $x^{3}+y^{3} = (x+y)(x^{2} -xy+y^{2} )$ )

we get,

= $(x+4) (x^{2} -x*4+4^{2} )$

= $(x+4)(x^{2} -4x+16)$ ....................(1)

solving the quadratic equation ,

$x^{2} -4x+16$ =0

solutions of this quadratic equation can be obtained by

$x=-b +- \sqrt{b^{2}-4ac } /2a$

let use y for factors

$x=-(-4x)+-\sqrt{(-4x^{2} )-4*x^{2} *16} / 2*x^{2}$

$x=4x+-\sqrt{16x^{2} -64x^{2} } /2x^{2}$

$x=4+-\sqrt{16-64}/2$

$x=4+-4\sqrt{3i} /2$

<u /> $x=2+-2\sqrt{3i}$ ..................(2)

from the equation 1 we have,

$x-4=0$

which gives solution $x=4$

and from equation 2 we got $x=2+-2\sqrt{3i}$

so the real solutions for the equation $x^{3}=-64$ are -

$x=4,2+-2\sqrt{3i}$

3 0

2 years ago