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3 years ago

Simplify (4x^3+13x-7) -(6x^2+9x+2)

Mathematics

1 answer:

svetoff [14.1K]3 years ago

8 0

$(4x^{3}+13x-7)-(6x^{2}+9x+2)\\\\=4x^{3}+13x-7-6x^{2}-9x-2\\\\=4x^{3}-6x^{2}+(13-9)x+(-7-2)\\\\=4x^{3}-6x^{2}+4x-9$

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Can anyone help me i was not a school and I need help

pashok25 [27]

so the right answer is of option A.

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3 0

3 years ago

Read 2 more answers

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marusya05 [52]

Roy is correct. When you multiply fractions, you multiply the numerator of one fraction and the other numerator of the other fraction, and vice versa for denominator.
7 x 3 = 21 (Numerators)
——- —-
3 x 8 = 32 (Denominators)

I hope this helped!

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3 years ago

Which equation has the solutions x = startfraction negative 3 plus-or-minus startroot 3 endroot i over 2 endfraction? 2x2 6x 9 =

Gennadij [26K]

The quadratic equation which matches given solution is x² + 3x + 3 = 0. Then the correct option is C.

<h3>What is a quadratic equation?</h3>

It is a polynomial that is equal to zero. Polynomial of variable power 2, 1, and 0 terms are there. Any equation having one term in which the power of the variable is a maximum of 2 then it is called a quadratic equation.

We know that the formula

$\rm x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

The solutions are given below.

$\rm x = \dfrac{-3 \pm \sqrt3 \iota }{2}$

Then

1) 2x² + 6x + 9 = 0, the zeroes of the equation will be

$\rm x = \dfrac{-6\pm 6\iota}{4}\\\\x = \dfrac{-3\pm 3 \iota}{2}$

2) x² + 3x + 12 = 0, the zeroes of the equation will be

$\rm x = \dfrac{-3\pm \sqrt{39}\iota}{2}$

3) x² + 3x + 3 = 0, the zeroes of the equation will be

$\rm x = \dfrac{-3\pm \sqrt3\iota}{2}$

4) 2x² + 6x + 3 = 0, the zeroes of the equation will be

$\rm x = \dfrac{-6\pm 2\sqrt{3}}{4}\\\\x = \dfrac{-3\pm \sqrt3 }{2}$

More about the quadratic equation link is given below.

brainly.com/question/2263981

4 0

2 years ago

A kite flier wondered how high her kite was flying. She used a protractor to measure an angle of 15∘ from level ground to the ki

Nataly [62]

Answer:

12.9 yd

Step-by-step explanation:

It helps if you draw a triangle. Draw a horizontal segment. That is the ground. Now at the left end start a new segment that goes up to the right at approximately 15 degrees until it its other endpoint directly above the right endpoint of the horizontal segment. Connect these two endpoints. The vertical side on the right shows the height of the kite. The hypotenuse is the string.

For the 15-deg angle, the height of the triangle is the opposite leg, and the string is the hypotenuse. The trig ratio that relates the opposite leg tot he hypotenuse is the sine.

$\sin A = \dfrac{opp}{hyp}$