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3 years ago

Solve for x:4=1/4(x-5)²

Mathematics

2 answers:

Paha777 [63]3 years ago

5 0

Answer:

x=9,1

Step-by-step explanation:

Take the root of both sides and solve.

x = 9 , 1

gladu [14]3 years ago

5 0

Answer:

X=9 or 1

Step-by-step explanation:

It's really complicated to explain, sorry!

(P.S, if that's you in your pfp, you are so freaking stunning, period. As you should)

Hope this helped, and please mark as brainliest :)

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Kamila [148]

Answer:

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Step-by-step explanation:

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2 years ago

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a cylinder has a base diameter of 16 cm and a height of 19 cmwhat is its volume in cubic centimeters?

kirza4 [7]

Answer:

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Step-by-step explanation:

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3 years ago

Heze has decided to save money for a new videogame He needs to have exactly $50 to purchase the game and has $5 already saved up

miv72 [106K]

Answer: 32.52

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3 years ago

How do I do 8b(ii) ? Please help me thank you!

Mila [183]

Step One
======
Find the length of FO (see below)

All of the triangles are equilateral triangles. Label the center as O
FO = FE = sqrt(5) + sqrt(2)

Step Two
======
Drop a perpendicular bisector from O to the midpoint of FE. Label the midpoint as J. Find OJ

Sure the Pythagorean Theorem. Remember that OJ is a perpendicular bisector.

FO^2 = FJ^2 + OJ^2
FO = sqrt(5) + sqrt(2)
FJ = 1/2 [(sqrt(5) + sqrt(2)] \
OJ = ??

[Sqrt(5) + sqrt(2)]^2 = [1/2(sqrt(5) + sqrt(2) ] ^2 + OJ^2
5 + 2 + 2*sqrt(10) = [1/4 (5 + 2 + 2*sqrt(10) + OJ^2
7 + 2sqrt(10) = 1/4 (7 + 2sqrt(10)) + OJ^2 Multiply through by 4
28 + 8* sqrt(10) = 7 + 2sqrt(10) + 4 OJ^2 Subtract 7 + 2sqrt From both sides
21 + 6 sqrt(10) = 4OJ^2 Divide both sides by 4
21/4 + 6/4* sqrt(10) = OJ^2
21/4 + 3/2 * sqrt(10) = OJ^2 Take the square root of both sides.
sqrt OJ^2 = sqrt(21/4 + 3/2 sqrt(10) )
OJ = sqrt(21/4 + 3/2 sqrt(10) )

Step three
find h
h = 2 * OJ
h = 2* sqrt(21/4 + 3/2 sqrt(10) ) <<<<<< answer.

7 0

3 years ago

Find the absolute maximum and minimum values of f on the set D. f(x,y)=2x^3+y^4, D={(x,y) | x^2+y^2<=1}.

castortr0y [4]

Answer:

absolute maximum is f(1, 0) = 2 and the absolute minimum is f(−1, 0) = −2.

Step-by-step explanation:

We compute,

$f_x = 6x^2, f_y=4y^3$

Hence, $f_x = f_y = 0$ if and only if (x,y) = (0,0)

This is unique critical point of D. The boundary equation is given by

$x^2+y^2=1$

Hence, the top half of the boundary is,

$$ T = \{ x, \sqrt{1-x^2} : -1 \leq x \leq 1\}$

On T we have, $f(x, \sqrt{1-x^2} = 2x^3 +(1-x^2)^2 = x^4 +2x^3-2x^2+1 \text{ for}\ -1 \leq x \leq 1$

We compute

$\frac{d}{dx}(f(x, \sqrt{1-x^2}))= 4x^3+6x^2-4x = 2x(2x^2+3x-2)=2x(2x-1)(x+2)=0$

0 if and only if x=0, x= 1/2 or x = -2.

We disregard $x = -2 \notin [-1,1]$

Hence, the critical points on T are (0,1) and $(\frac{1}{2}, \sqrt{1-(\frac{1}{2})^2}=\frac{\sqrt3}{2})$

On the bottom half, B, we have

$f(x, \sqrt{1-x^2})= f(x,-\sqrt{1-x^2})$

Therefore, the critical points on B are (0,-1) and $$( 1/2, -\sqrt3/2)$

It remains to evaluate f(x, y) at the points $(0,0), (0 \pm1), (1/2, \pm \sqrt3/2) \text{ and}\ (\pm1, 0)$ .

We should consider latter two points, $(\pm1,0)$ , since they are the boundary points for the T and also B. We compute $f(0,0)=0, \ \f(0 \pm1)=1, \ \ f(0, \pm \sqrt3/2)=9/16, \ \ f(1,0 )= 2 \text{ and}\ \ f(-1,0)= -2$

We conclude that the absolute maximum = f(1, 0) = 2

And the absolute minimum = f(−1, 0) = −2.

6 0

3 years ago

Solve for x:4=1/4(x-5)²​

Solve for x:4=1/4(x-5)²