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vivado [14]
3 years ago
13

Can someone help me with this?

Mathematics
1 answer:
Ann [662]3 years ago
3 0
So first you have to add all of the sides and then you have to multiply 
                          Hope it helped 
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Helppppp and show equation plsssssssss
laiz [17]

Answer:

24

Step-by-step explanation:

6+2×3+4+(27-19)

first the bracket, (27 -19) = 8

and then the multiplication, 2×3 = 6

then the addition, 6+6+4+8 = 24

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Sindrei [870]
A quantity in a standard time or period maybe lol
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3 years ago
My teacher split c pieces of candy among 23 students.<br> Writing Expressions with answers please.
Kryger [21]

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Step-by-step explanation:

8 0
3 years ago
Find the distance from the origin to the graph of 7x+9y+11=0
Cerrena [4.2K]
One way to do it is with calculus. The distance between any point (x,y)=\left(x,-\dfrac{7x+11}9\right) on the line to the origin is given by

d(x)=\sqrt{x^2+\left(-\dfrac{7x+11}9\right)^2}=\dfrac{\sqrt{130x^2+154x+121}}9

Now, both d(x) and d(x)^2 attain their respective extrema at the same critical points, so we can work with the latter and apply the derivative test to that.

d(x)^2=\dfrac{130x^2+154x+121}{81}\implies\dfrac{\mathrm dd(x)^2}{\mathrm dx}=\dfrac{260}{81}x+\dfrac{154}{81}

Solving for (d(x)^2)'=0, you find a critical point of x=-\dfrac{77}{130}.

Next, check the concavity of the squared distance to verify that a minimum occurs at this value. If the second derivative is positive, then the critical point is the site of a minimum.

You have

\dfrac{\mathrm d^2d(x)^2}{\mathrm dx^2}=\dfrac{260}{81}>0

so indeed, a minimum occurs at x=-\dfrac{77}{130}.

The minimum distance is then

d\left(-\dfrac{77}{130}\right)=\dfrac{11}{\sqrt{130}}
4 0
3 years ago
8 hours (hrs) = minutes (min)
LenKa [72]

Answer: 480 minutes

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6 0
3 years ago
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