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Ira Lisetskai [31]
3 years ago
7

Three consecutive even integers have a sum of 18. find the integers ​

Mathematics
1 answer:
natta225 [31]3 years ago
5 0

Answer:

5,6,7

Step-by-step explanation:

x-1+x+x+1=18

3x=18

x=6

5,6,7

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The volume of a cube-shaped box is 27^5 cubic millimeters. What's the length of one side of the cube-shaped box?
Vinil7 [7]
5^10 hope this helps have a good day
7 0
3 years ago
What is the value of x?<br> Enter your answer in the box.<br><br> x= ____
Nady [450]

x^2 + 6^2 = 10^2

x^2 + 36 = 100

x^2 = 100-36

x^2 = 64

x = sqrt(64)

x = 8

3 0
3 years ago
Read 2 more answers
Plz help me fast Simplify t12/t6
Digiron [165]

Answer:

t^6

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when dividing exponents, you subtract (refer to exponent rule)

t^12 - t^6 = t^6

4 0
3 years ago
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J is the midpoint of HK¯¯¯¯¯¯¯ . What are HJ, JK, and HK?
lana [24]
Since J is the midpoint of HK, that means HK is split into two sections HJ and JK that are the same length.

1) You are told that the m<span>easure of segment HJ = 9x-2 and that of segment JK = 4x+13. Since you also know they are equal lengths, you can set these equations equal to each other to find the value of x!
HJ = JK
</span>9x-2 = 4x+13
5x = 15
x = 3

2) Now you know x = 3. Plug that into your given equations for HJ and JK to find the length of each segment (or a shortcut would be to find one of them, and then you also know the other is the same length. I'm doing both, just to make sure I don't make a silly mistake!):
HJ = <span>9x-2 
</span>HJ = 9(3) - 2
HJ = 27 - 2
HJ = 25

JK = 4x + 13
JK = 4(3) + 13
JK = 12 + 13
JK = 25

3) Finally, the length of HK is just the length of HJ + JK, or HK = 25 + 25 = 50.

-----

Answer: HJ = 25, JK = 25, HK = 50
4 0
3 years ago
Find the maximum and minimum values of the function f(x,y)=2x2+3y2−4x−5 on the domain x2+y2≤100. The maximum value of f(x,y) is:
ryzh [129]

First find the critical points of <em>f</em> :

f(x,y)=2x^2+3y^2-4x-5=2(x-1)^2+3y^2-7

\dfrac{\partial f}{\partial x}=2(x-1)=0\implies x=1

\dfrac{\partial f}{\partial y}=6y=0\implies y=0

so the point (1, 0) is the only critical point, at which we have

f(1,0)=-7

Next check for critical points along the boundary, which can be found by converting to polar coordinates:

f(x,y)=f(10\cos t,10\sin t)=g(t)=295-40\cos t-100\cos^2t

Find the critical points of <em>g</em> :

\dfrac{\mathrm dg}{\mathrm dt}=40\sin t+200\sin t\cos t=40\sin t(1+5\cos t)=0

\implies\sin t=0\text{ OR }1+5\cos t=0

\implies t=n\pi\text{ OR }t=\cos^{-1}\left(-\dfrac15\right)+2n\pi\text{ OR }t=-\cos^{-1}\left(-\dfrac15\right)+2n\pi

where <em>n</em> is any integer. We get 4 critical points in the interval [0, 2π) at

t=0\implies f(10,0)=155

t=\cos^{-1}\left(-\dfrac15\right)\implies f(-2,4\sqrt6)=299

t=\pi\implies f(-10,0)=235

t=2\pi-\cos^{-1}\left(-\dfrac15\right)\implies f(-2,-4\sqrt6)=299

So <em>f</em> has a minimum of -7 and a maximum of 299.

4 0
3 years ago
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