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frosja888 [35]
2 years ago
5

5 Anita is using 36 cards to set up a card game. She wants to make

Mathematics
1 answer:
serious [3.7K]2 years ago
3 0

Answer:

box is there so I'll put the money in the post

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Find the other endpoint of the line segment with the given endpoint and midpoint.
Shtirlitz [24]

Answer:

endpoint is (29, -13)

Step-by-step explanation:

Formula for midpoint is (\frac{x_{1} + x_{2} }{2} ,\frac{y_{1} + y_{2}  }{2} )

So, let (x_{2} , y_{2} ) be the coordinates of the endpoint you are looking for.

x_{1} = -9  and y_{1}  = 7

(\frac{-9 + x_{2} }{2} , \frac{7 + y_{2} }{2} ) = (10, -3)

\frac{-9 + x_{2} }{2}  = 10           \frac{7 + y_{2} }{2}  = -3

-9 + x_{2} = 20         7 + y_{2}  = -6

x_{2}  =  29                    y_{2}  = -13

endpoint is (29, -13)

6 0
3 years ago
Please help me asap! please show your work
solmaris [256]
Make a proportion
81/30 = x/600
600/30 = 20
x/81 = 20
x = 1620
Solution: 1620 calories
7 0
3 years ago
Gina bought 2.3 pounds of red apples and 2.42 pounds of green apples. They were on sale for $0.75 a pound. How much did the appl
vampirchik [111]
4. 72•0.75= $3.54 is the correct answer
3 0
3 years ago
There are two numbers. Their GCF is 5. Their LCM is 150. The first number is 25. What is the second number?
Galina-37 [17]

Answer:

6

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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
4 years ago
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