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jeka57 [31]
3 years ago
13

Which runner came in first place 8.016 or 8.16

Mathematics
1 answer:
antiseptic1488 [7]3 years ago
6 0
8.16 = 8.160
So, 8.016 < 8.16
As, 8.016 is smaller, so it will come first.
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Someome please help me
MrRa [10]

Answer:

Step-by-step explanation:

Tell your teacher thet you need more helpp and pay attention to math

3 0
3 years ago
Quick I need help 23 and 24 easy seventh grade math
Alekssandra [29.7K]

Answer:-14, -26

Step-by-step explanation:

13-4=9

4-9=-5

From here you can see that the pattern is -9

-5-9=-14

23. is -14

16-2=14

2-14=-12

From here you can see that the pattern is -14

-12-14=-26

24. is -14

6 0
3 years ago
Find the greatest solution for x+y when x^2+y^2 = 7, x^3+y^3=10
damaskus [11]

Answer:

4

Step-by-step explanation:

set

f(x,y)=x+y\\

constrain:

g(x,y)=x^2+y^2 = 7\\h(x,y)=x^3+y^3=10

Partial derivatives:

f_{x}=1\\f_{y} =1 \\g_{x}=2x \\g_{y}=2y\\h_{x}=3x^2 \\h_{y}=3y^2

Lagrange multiplier:

grad(f)=a*grad(g)+b*grad(h)\\

\left[\begin{array}{ccc}1\\1\end{array}\right]=a\left[\begin{array}{ccc}2x\\2y\end{array}\right]+b\left[\begin{array}{ccc}3x^2\\3y^2\end{array}\right]

4 equations:

1=2ax+3bx^2\\1=2ay+3by^2\\x^2+y^2=7\\x^3+y^3=10

By solving:

a=4/9\\b=-2/27\\x+y=4

Second mathod:

Solve for x^2+y^2 = 7, x^3+y^3=10 first:

x=\frac{1}{2} -\frac{\sqrt{13}}{2} \ or \ y=\frac{1}{2} +\frac{\sqrt{13}}{2} \\x=\frac{1}{2} +\frac{\sqrt{13}}{2} \ or \ y=\frac{1}{2} -\frac{\sqrt{13}}{2} \\x+y=-5\ or\ 1 \or\ 4

The maximum is 4

6 0
3 years ago
Which quotient matches the description of solving with partial quotients for 240÷15
Illusion [34]

Answer:

C. 16

Step-by-step explanation:

240/15

Step one: 10x15= 150

              240-150= 90  

Partial quotient: 10

Step two: 6x15= 90

                 90-90= 0

Partial quotient: 6

Step 3: 10+6= 16

Answer check: 240/15= 16

3 0
3 years ago
Vanessa started with $800 in a bank account that does not earn interest. In the middle of every month, she withdraws 14 of the a
Nataliya [291]
An =800 *an-1 an a1=14
3 0
3 years ago
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