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sp2606 [1]
3 years ago
14

Which of the following is NOT a perfect square trinomial?

Mathematics
1 answer:
DIA [1.3K]3 years ago
7 0
The second one is a perfect square trinomial
X^2+16x+64
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Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.
lbvjy [14]

Answer:

<em>π∫52(ey3)2dy</em>

The work I did to solve this equation:

Step 1

<em>ln(3x)=2 </em>

<em>3x=2e </em>

<em>x=2e3 </em>

Step 2

<em>ln(3x)=5 </em>

<em>3x=5e </em>

<em>x=5e3</em>

Step 3

<em>y=ln(3x)⟺ey=3x⟺ey3=x</em>

Step 4

π∫52(ey3)2dy

3 0
3 years ago
I WILL GIVE 20 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS PLEASE
kotykmax [81]
The answer is 1,2,4,6,8 because the circle has a circumference of 9
4 0
2 years ago
Read 2 more answers
1+-w2+9w and I need help cuz I’m on 76 and I’m sooo close help
Gnesinka [82]

\huge \boxed{\mathfrak{Question} \downarrow}

  • Simplify :- 1 + - w² + 9w.

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

\large \sf1 + - w ^ { 2 } + 9 w

Quadratic polynomial can be factored using the transformation \sf \: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where \sf x_{1} and x_{2} are the solutions of the quadratic equation \sf \: ax^{2}+bx+c=0.

\large \sf-w^{2}+9w+1=0

All equations of the form \sf\:ax^{2}+bx+c=0 can be solved using the quadratic formula: \sf\frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

\large \sf \: w=\frac{-9±\sqrt{9^{2}-4\left(-1\right)}}{2\left(-1\right)}  \\

Square 9.

\large \sf \: w=\frac{-9±\sqrt{81-4\left(-1\right)}}{2\left(-1\right)}  \\

Multiply -4 times -1.

\large \sf \: w=\frac{-9±\sqrt{81+4}}{2\left(-1\right)}  \\

Add 81 to 4.

\large \sf \: w=\frac{-9±\sqrt{85}}{2\left(-1\right)}  \\

Multiply 2 times -1.

\large \sf \: w=\frac{-9±\sqrt{85}}{-2}  \\

Now solve the equation \sf\:w=\frac{-9±\sqrt{85}}{-2} when ± is plus. Add -9 to \sf\sqrt{85}.

\large \sf \: w=\frac{\sqrt{85}-9}{-2}  \\

Divide -9+ \sf\sqrt{85} by -2.

\large \boxed{ \sf \: w=\frac{9-\sqrt{85}}{2}} \\

Now solve the equation \sf\:w=\frac{-9±\sqrt{85}}{-2} when ± is minus. Subtract \sf\sqrt{85} from -9.

\large \sf \: w=\frac{-\sqrt{85}-9}{-2}  \\

Divide \sf-9-\sqrt{85} by -2.

\large \boxed{ \sf \: w=\frac{\sqrt{85}+9}{2}}  \\

Factor the original expression using \sf\:ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \sf\frac{9-\sqrt{85}}{2}for \sf\:x_{1} and \sf\frac{9+\sqrt{85}}{2} for \sf\:x_{2}.

\large \boxed{ \boxed {\mathfrak{-w^{2}+9w+1=-\left(w-\frac{9-\sqrt{85}}{2}\right)\left(w-\frac{\sqrt{85}+9}{2}\right) }}}

<h3>NOTE :-</h3>

Well, in the picture you inserted it says that it's 8th grade mathematics. So, I'm not sure if you have learned simplification with the help of biquadratic formula. So, if you want the answer simplified only according to like terms then your answer will be ⇨

\large \sf \: 1 + -  w {}^{2}  + 9w \\  =\large  \boxed{\bf \: 1 -  {w}^{2}   + 9w}

This cannot be further simplified as there are no more like terms (you can use the biquadratic formula if you've learned it.)

4 0
2 years ago
The given lengths are two sides of a right triangle. All three side lengths of the triangle are integers and together form a pyt
tatuchka [14]
24^2 + 45^2 = 51^2
576 + 2025 = 2601
2601 = 2601

The answer is B leg.
3 0
3 years ago
Read 2 more answers
Write the slope-intercept form of the equation of each line
viva [34]
Answers
1. y=9/7x +1
2. y= 11/4x -8
3. y= 11/8x +6
See work in attachment

4 0
2 years ago
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