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Harlamova29_29 [7]
3 years ago
5

Help me please I'll give you brainliest ​

Mathematics
1 answer:
Ber [7]3 years ago
5 0

Answer:

x means

Step-by-step explanation:

10x means that the x is something you would times to get the number you want which is 44 so 10 times x=4. hope this helped :)

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I don't know give me the best joke and i mark you brainliest :)
Greeley [361]

Answer:

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Write an equation that represents the following word problem. Jackie has four more apples than Mark and together they have twelv
Vesnalui [34]

Answer:

also jackie has 8 apples and mark has 4 apples

option B is correct

Step-by-step explanation:

a+4=b =>

a+b=12

(a+4) + a =12

4 0
3 years ago
What is the solution to<br> 4(2x– 3)= 2(3x+ 1)?<br><br><br> -5<br><br> 1<br><br> 7<br><br> 10
Paul [167]
4 (2x – 3)= 2( 3x + 1)
<span>(4 (2x – 3)= 2( 3x + 1)) </span>÷ 2
2 (2x – 3) = (3x + 1)
4x - 6 = 3x +1
-3x       -3x
__________
x - 6 = 1
   +6  +6
______
x = 7
8 0
3 years ago
What are the vertical asymptotes of the function f(x) =5x+5/x2 + x-2
katen-ka-za [31]

let's recall that the vertical asymptotes for a rational expression occur when the denominator is at 0, so let's zero out this one and check.

\bf \cfrac{5x+5}{x^2+x-2}\qquad \stackrel{\textit{zeroing out the denominator}~\hfill }{x^2+x-2=0\implies (x+2)(x-1)=0}\implies \stackrel{\textit{vertical asymptotes}}{ \begin{cases} x=-2\\ x=1 \end{cases}}

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3 years ago
Use a surface integral to find the surface area of the portion of the sphere xUse a surface integral to find the surface area of t
brilliants [131]

Parameterize this surface (call it S) by

\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos v\,\vec k

with 0\le u\le2\pi and 0\le v\le\dfrac\pi4. The limits on u should be obvious. We find the upper limit for v by solving for v where the sphere and cone intersect:

\begin{cases}x^2+y^2+z^2=1\\z=\sqrt{x^2+y^2}\end{cases}\implies x^2+y^2=\dfrac12

\implies(\cos u\sin v)^2+(\sin u\sin v)^2=\dfrac12

\implies\sin^2v=\dfrac12

\implies\sin v=\dfrac1{\sqrt2}\implies v=\dfrac\pi4

Take the normal vector to S to be

\vec r_u\times\vec r_v=-\cos u\sin^2v\,\vec\imath-\sin u\sin^2v\,\vec\jmath-\cos v\sin v\,\vec k

(orientation does not matter here)

Then the area of S is

\displaystyle\iint_S\mathrm dA=\iint_S\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^{\pi/4}\int_0^{2\pi}\sin v\,\mathrm du\,\mathrm dv

=\displaystyle2\pi\int_0^{\pi/4}\sin v\,\mathrm dv=\boxed{(2-\sqrt2)\pi}

4 0
3 years ago
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