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2 years ago

I really had some trouble with this problem:( & I need some help

Mathematics

1 answer:

Natasha_Volkova [10]2 years ago

7 0

The answers are in the photo

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If x²-5x +1 =0, what is the value of -<br><img src="https://tex.z-dn.net/?f=%28x%20%2B%20%20%5Cfrac%7B1%7D%7Bx%7D%20%29%20%20" i

ra1l [238]

<h3>Answer: 5</h3>

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Work Shown:

x^2 - 5x + 1 = 0

x^2 + 1 - 5x = 0

x^2 + 1 = 5x

(x^2 + 1)/x = 5 .... where x is nonzero

(x^2)/x + (1/x) = 5

x + (1/x) = 5

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An alternative method involves solving the original equation using the quadratic formula. After you get the two roots x = p and x = q, you should be able to find that p + 1/p = 5 and also q + 1/q = 5 as well.

In this case,

p = (5 + sqrt(21))/2

q = (5 - sqrt(21))/2

8 0

3 years ago

Read 2 more answers

ST has endpoints Y(-2, -4) and T (6, 8). Find the coordinates of the midpoint of ST

storchak [24]

add up the X values and y values and divide by two

X=(-2+6)/2=2

Y=(-4=8)/2=2

5 0

3 years ago

Read 2 more answers

Out of 30 questions, Ahmad answer 12 of them incorrectly. What percent of the questions did he answer correctly?

kati45 [8]

First, you would need to determine how many questions he got correct.
30 - 12 = 18

Next, you would need to create an equation for this problem. Let x represent the unknown percentage.
$\frac{18}{30}$ = $\frac{x}{100}$

Now, you would cross multiply.
30x = 1,800

The last step would be to isolate the x. To do this, you would divide both sides of the equal sign by 30.
x = 60

Ahmad got 60% of the questions correct.

I hope this helps!

7 0

3 years ago

Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )

DedPeter [7]

(a) First find the intersections of $y=e^{2x-x^2}$ and $y=2$ :

$2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}$

So the area of $R$ is given by

$\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx$

If you're not familiar with the error function $\mathrm{erf}(x)$ , then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line $y=1$ with $y=e^{2x-x^2}$ .

$1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2$

So the area of $S$ is given by

$\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx$
$\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx$

which is approximately 1.546.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve $y=e^{2x-x^2}$ and the line $y=1$ , or $e^{2x-x^2}-1$ . The area of any such circle is $\pi$ times the square of its radius. Since the curve intersects the axis of revolution at $x=0$ and $x=2$ , the volume would be given by

$\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx$

5 0

3 years ago

What is the remainder of (4x2 + 7x-1)= (4 + x)?<br>A. -9x – 1<br>B.23x – 1<br>C.35<br>D.-37

bearhunter [10]

Answer: I don't know what you meant by remainder but i hope this helps :)

$x=\frac{-3+\sqrt{29}}{4},\:x=-\frac{3+\sqrt{29}}{4}\\$

Step-by-step explanation:

$\left(4x^2+7x-1\right)=\left(4+x\right)\\\mathrm{Refine}\\4x^2+7x-1=4+x\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\4x^2+7x-1-x=4+x-x\\Simplify\\4x^2+6x-1=4\\\mathrm{Subtract\:}4\mathrm{\:from\:both\:sides}\\4x^2+6x-1-4=4-4\\\mathrm{Simplify}\\4x^2+6x-5=0\\\mathrm{For\:}\quad a=4,\:b=6,\:c=-5:\\\quad x_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\times \:4\left(-5\right)}}{2\times \:4}$

$\frac{-6+\sqrt{6^2-4\times \:4\left(-5\right)}}{2\times \:4}\\=\frac{-6+\sqrt{6^2+4\times \:4\times \:5}}{2\times \:4}\\=\frac{-6+\sqrt{116}}{2\times \:4}\\=\frac{-6+\sqrt{116}}{8}\\\\Let\: simplify\: ; -6+2\sqrt{29}\\=-2\times \:3+2\sqrt{29}\\=2\left(-3+\sqrt{29}\right)\\=\frac{2\left(-3+\sqrt{29}\right)}{8}\\=\frac{-3+\sqrt{29}}{4}\\$

$\frac{-6-\sqrt{6^2-4\times \:4\left(-5\right)}}{2\times \:4}\\\\=\frac{-6-\sqrt{6^2+4\times \:4\times \:5}}{2\times \:4}\\\\=\frac{-6-\sqrt{116}}{2\times \:4}\\\\=\frac{-6-2\sqrt{29}}{8}\\\\=-\frac{2\left(3+\sqrt{29}\right)}{8}\\\\=-\frac{3+\sqrt{29}}{4}\\\\\\x=\frac{-3+\sqrt{29}}{4},\:x=-\frac{3+\sqrt{29}}{4}$

6 0

3 years ago

Read 2 more answers