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2 years ago

1. Use what you have learned to find the solutions of the polynomial equation shown below.

1 answer:

4 0

If (x^2 -10) is one of the factors, that can be further factored into:
(x - sqrt(10) ) * (x+sqrt(10)) =0
making 2 of the 4 solutions equal:
3.1623 and -3.1623

I then used an algebraic long division calculator
http://calculus-calculator.com/longdivision/
to calculate:
<span>x^4 + 5x^3 ‒ x^2 ‒ 50x ‒ 90 divided by x^2 -10 which equals

</span>x^2 + 5x + 9

Using the quadratic formula, the roots of that equation are:
x = -5 + sqrt (-11) / 2
and
x = -5 - sqrt (-11) / 2

Both of those roots are not real.

I tried using online graphing calculators for x^4+5x^3-x^2-50x-90=0 but none worked.

2. For this equation,
<span>3x^2 ‒ 8x + k = 0
I used my OWN quadratic formula calculator
http://www.1728.org/quadratc.htm
and found that real roots no longer exist after "k" is greater than 5.3

</span>

You might be interested in

X + 2y = 7<br> 3x - 2y = -3

VladimirAG [237]

Use Subtraction

1. Subtract
x+2y=7
3x-2y=-3
-------------
4x+0y=4 SIMPLIFY 4x=4

2. DIVIDE

4x=4
x=1

Substitute that into either of the original equations to get x=1, y=3

4 0

3 years ago

If VT=2,what is the length of PT?

marusya05 [52]

Step-by-step explanation:

$\underline{ \underline{ \text{Given : }}}$

Perpendicular ( P ) = VT = 2
Base ( b ) = PT
$\theta = \tt{60 \degree}$

$\underline{ \underline{ \text{Solution}}} :$

$\tt{ \tan(60 \degree) = \frac{perpendicualar}{base}}$

⟶ $\tt{ \sqrt{3} = \frac{2}{PT}}$

⟶ $\tt{ \sqrt{3} \: PT= 2}$

⟶ $\tt{PT = \frac{2}{ \sqrt{3}} }$

⟶ $\tt{PT= \boxed{\tt{ \frac{2 \sqrt{3} }{{3} }}}}$

Hope I helped ! ♡

Have a wonderful day / night ! ツ

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6 0

3 years ago

Is this correct? formula is displayed on image:

Len [333]

Answer:

no it should bee

$\tan( \alpha ) = \frac{15}{8}$

7 0

2 years ago

Read 2 more answers

Use the identity below to complete the tasks:

Gennadij [26K]

Answer:

a3+b3=(a+b)(a2-ab+b2)

4 0

3 years ago

X²+Y²=250 and XY=117<br> What are the values of X and Y?

yan [13]

Answer:

Step-by-step explanation:

$x^2+y^2=250\\\\x^2-2xy+y^2+2xy=250\\\\(x-y)^2=250-2xy\\\\(x-y)^2=250-2\cdot117\\\\ (x-y)^2=16\\\\x-y=4\qquad\qquad\vee\qquad \qquad x-y=-4\\\\x=4+y \qquad\qquad \vee\qquad\qquad x=-4+y\\\\(y+4)y=117\qquad\vee\qquad\quad (y-4)y=117\\\\y^2+4y-117=0\qquad\vee\qquad y^2-4y-117=0\\\\y=\dfrac{-4\pm\sqrt{4^2-4(-117)}}{2\cdot1}\qquad\vee\qquad y=\dfrac{4\pm\sqrt{4^2-4(-117)}}{2\cdot1}\\\\y=\dfrac{-4\pm\sqrt{16+468}}{2}\qquad\ \ \vee\qquad y=\dfrac{4\pm\sqrt{16+468}}{2}$

$y_1=\dfrac{-4-22}{2}\ ,\quad y_2=\dfrac{-4+22}{2}\ ,\quad y_3=\dfrac{4-22}{2}\ ,\quad y_4=\dfrac{4+22}{2}\\\\y_1=-13\ ,\qquad y_2=9\ ,\qquad\quad\qquad\ y_3=-9\ ,\qquad y_4=13\\\\x_{1,2}=4+y_{1,2}\qquad\qquad\qquad\qquad\qquad x_{3,4}=-4+y_{3,4}\\\\x_1=-9\ ,\qquad x_2=13\ ,\qquad\quad\qquad x_3=-13\ ,\qquad x_4=9$

6 0

3 years ago

1. ​​Use what you have learned to find the solutions of the polynomial equation shown below.

1. Use what you have learned to find the solutions of the polynomial equation shown below.