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3 years ago

Solve the square: x^2-4x+8=0

Mathematics

1 answer:

skelet666 [1.2K]3 years ago

4 0

Answer:

(x-2)^2+4.

Hope this helps!

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What to the tenth power can get you 300

OLEGan [10]

Answer:

1.7689 (rounded to 4 decimal places)

Step-by-step explanation:

Let the number we are seeking be "x", thus we can write the equation as:

$x^{10}=300$

Since we have raised "x" to the "10th power", to get "x" back again, we need to take 10th root. Same goes for right side, we take 10th root of 300. We will get our answer. The process shown below:

$x^{10}=300\\\sqrt[10]{x^{10}} =\sqrt[10]{300} \\x=\sqrt[10]{300} \\x=1.7689$

hence, 1.7689 raised to 10th power will give us 300

5 0

3 years ago

General solutions of sin(x-90)+cos(x+270)=-1<br> {both 90 and 270 are in degrees}

mixer [17]

Answer:

$\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

Step-by-step explanation:

Given:

$\sin (x-90^{\circ})+\cos(x+270^{\circ})=-1$

First, note that

$\sin (x-90^{\circ})=-\cos x\\ \\\cos(x+270^{\circ})=\sin x$

So, the equation is

$-\cos x+\sin x= -1$

Multiply this equation by $\frac{\sqrt{2}}{2}:$

$-\dfrac{\sqrt{2}}{2}\cos x+\dfrac{\sqrt{2}}{2}\sin x= -\dfrac{\sqrt{2}}{2}\\ \\\dfrac{\sqrt{2}}{2}\cos x-\dfrac{\sqrt{2}}{2}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos (x+45^{\circ})=\dfrac{\sqrt{2}}{2}$

The general solution is

$x+45^{\circ}=\pm \arccos \left(\dfrac{\sqrt{2}}{2}\right)+2\pi k,\ \ k\in Z\\ \\x+\dfrac{\pi }{4}=\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{4}\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

4 0

3 years ago

EXTRA POINTS

hram777 [196]

Answer:

Step-by-step explanation:

The volume of the pyramid = (1/3)*area of base *height

= (1/3)*10*24*13 = 1040 cubic units.

The total surface area = area of rectangular base + area of 2 isosceles triangles with a base of 24 units + area of 2 isosceles triangles with a base of 10 units.

Area of rectangular base = 24*10 = 240 sq units.

The slant height of isosceles triangles with a base of 24 units = [(10/2)^2+13^2]^0.5 = [25+169]^0.5 = 194^0.5 = 13.92838828 units.

The area of 2 isosceles triangles with a base of 24 units 2*24*13.92838828/2 = 334.2813187 sq units.

The slant height of isosceles triangles with a base of 10 units = [(24/2)^2+13^2]^0.5 = [144+169]^0.5 = 194^0.5 = 17.69180601 units.

The area of 2 isosceles triangles with a base of 10 units 2*10*17.69180601/2 = 176.9180601 sq units.

The total surface area of the pyramid = 240 + 334.2813187 + 176.9180601 = 591.9731247 sq units.

8 0

3 years ago

Read 2 more answers

Ebony walked at arate of 3 1/2 miles per hour for 1 1/3 how far did she walk?

ElenaW [278]

let's firstly convert the mixed fractions to improper fractions.

$\bf \stackrel{mixed}{3\frac{1}{2}}\implies \cfrac{3\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{7}{2}}~\hfill \stackrel{mixed}{1\frac{1}{3}}\implies \cfrac{1\cdot 3+1}{3}\implies \stackrel{improper}{\cfrac{4}{3}} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \begin{array}{ccll} miles&hours\\ \cline{1-2}\\ \frac{7}{2}&1\\[0.8em] m&\frac{4}{3} \end{array}\implies \cfrac{~~\frac{7}{2}~~}{m}=\cfrac{~~1~~}{\frac{4}{3}}\implies \cfrac{7}{2m}=\cfrac{3}{4}\implies 28=6m \\\\\\ \cfrac{28}{6}=m\implies \cfrac{14}{3}=m\implies 4\frac{2}{3}=m$

6 0

3 years ago

Factor the following expression.<br> 6x^2- 5x-1

jenyasd209 [6]

Answer:

X=(6x+1)(x-1) because you are factoring out

8 0

3 years ago