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Allushta [10]
2 years ago
15

Kyle has 105 baseball cards and must give 1/5 of his figures away. How many cards will Kyle give away? What fraction of his card

s does he have left?​
Mathematics
1 answer:
andreev551 [17]2 years ago
4 0

Answer:

I think it would be 21, because 105 / 5 is 21, and if the denominator is the holder spot, it should be 21

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PLEASE HELP I HAVE 3 MINS LEFT
lyudmila [28]
I believe it is 5 or 10
5 0
2 years ago
The scale below shows the weight of one apple. If Carly put another apple on the scale with the same weight as the first apple,
lara [203]

Answer:

10

Step-by-step explanation:

if you add 5.5 + 5.5 together its 10

8 0
2 years ago
there are 36 roses and 27 carnations. anna is making flower arrangements using both flowers, if she made the maximum number of a
solniwko [45]
This is a GCF problem,
find the GCF of 36 and 27. To do this, list out the factors for each number.
EX 36- 1×36, 2×18, so on and so on. Then do this for 27. The GCF will be the greatest factor. In this case, that is 9.
So there would be 9 groups because that is the greatest common factor.
there would be 4 roses in each group because 4×9= 36.
There are 3 carnations in each group because 9×3 =27.
So 9 groups with 4 roses and 3 carnations.

4 0
3 years ago
Consider the parabola r​(t)equalsleft angle at squared plus 1 comma t right angle​, for minusinfinityless thantless thaninfinity
kodGreya [7K]

Given:-   r(t)=< at^2+1,t>  ; -\infty < t< \infty , where a is any positive real number.

Consider the helix parabolic equation :  

                                              r(t)=< at^2+1,t>

now, take the derivatives we get;

                                            r{}'(t)=

As, we know that two vectors are orthogonal if their dot product is zero.

Here,  r(t) and r{}'(t)  are orthogonal i.e,   r\cdot r{}'=0

Therefore, we have ,

                                  < at^2+1,t>\cdot < 2at,1>=0

< at^2+1,t>\cdot < 2at,1>=

                                              =2a^2t^3+2at+t

2a^2t^3+2at+t=0

take t common in above equation we get,

t\cdot \left (2a^2t^2+2a+1\right )=0

⇒t=0 or 2a^2t^2+2a+1=0

To find the solution for t;

take 2a^2t^2+2a+1=0

The numberD = b^2 -4ac determined from the coefficients of the equation ax^2 + bx + c = 0.

The determinant D=0-4(2a^2)(2a+1)=-8a^2\cdot(2a+1)

Since, for any positive value of a determinant is negative.

Therefore, there is no solution.

The only solution, we have t=0.

Hence, we have only one points on the parabola  r(t)=< at^2+1,t> i.e <1,0>




                                               




6 0
3 years ago
Lexi bought 1 2/5 of green grapes and 2 3/10 of red grapes. Grapes are 2.99 a pound.Did lexi spend more or less then $12?
Romashka-Z-Leto [24]
Less. If you multiple 1 2/5 by $2.99 you would get about $4.19. If you multiple 2 3/10 by $2.99 you would get about $6.88. If you add those together, you would get $11.07, which is less than $12.
4 0
3 years ago
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