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Alchen [17]
3 years ago
13

T (10 x k) solve if t=2 and k=6 show work please

Mathematics
1 answer:
grigory [225]3 years ago
5 0

Answer:

120

Step-by-step explanation:

If t=2 and k=6 then this is what you do:

- 2(10*6)

- 10*6=60

- 2*60=120

HOPE THIS HELPS!

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if we increased one side of a square by 5 units and decreased the other by 3 units the area of the resulting rectangle would be
zvonat [6]

Answer:

18 units

Step-by-step explanation:

So let's list out the sides.

for the first square let's just call them x

for the second square then they would be x+5 and x-3

So let's write out their areas we will cal the area of the first one z

x*x = z

(x+5)*(x-3) = z+21

since z = x^2 we can set up the second equation as a quadratic.

(x+5)*(x-3) = x^2 + 21

x^2 - 3x + 5x - 15 = x^2 + 21

But look, the x^2s cancel out

2x - 15 = 21

2x = 36

x = 18

Test it out and see if it fits the description, And if you don't understand anything just let me know so I can explain more.  

3 0
3 years ago
How can you write 1,360/200 as a repeating decimal?
Alisiya [41]
1360/200

13.6/2

6.8

It is a finite decimal, but if you want repeating, there are zeroes that follow, like so:

6.8000000000000 . . . (and on to infinity)
6 0
3 years ago
14. Let R^2 have inner product defined by ((x1,x2), (y,, y2)) 4x1y1 +9x2y2 A. Determine the norm of (-1,2) in this space B. Dete
Vitek1552 [10]

The norm of a vector \vec x is equal to the square root of the inner product of \vec x with itself.

a. \|(-1,2)\|=\sqrt{\langle(-1,2),(-1,2)\rangle}=\sqrt{4(-1)^2+9(2)^2}=\sqrt{40}=2\sqrt{10}

b. \|(3,2)\|=\sqrt{\langle(3,2),(3,2)\rangle}=\sqrt{4(3)^2+9(2)^2}=\sqrt{72}=6\sqrt2

7 0
4 years ago
What is the standard equation with center (4, -5) and radius 11
Marina86 [1]

Answer: (x-4)² +(y+5)²=121

Step-by-step explanation:

Happy to help :)

3 0
3 years ago
The second deposit that apex savings bank received was $12,700. If apex savings bank was required to keep $571.50 on reserve, wh
vodomira [7]


12700 \div 571.50 = 22.22
4 0
4 years ago
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