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shepuryov [24]
3 years ago
10

Which is longer 1/3 of a minute or 2/3 of a minute?

Mathematics
2 answers:
damaskus [11]3 years ago
8 0
2/3 of a minute, because 2/3 is greater than 1/3.
Evgesh-ka [11]3 years ago
3 0
2/3 of a minute is longer than 1/3 of a minute
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What is the value for a?
alexgriva [62]
The value of a is 5
6 0
2 years ago
£980 is divided between Caroline, Sarah & Gavyn so that Caroline gets twice as much as Sarah, and Sarah gets three times as
Nitella [24]

Sarah has received £ 294

<u><em>Solution:</em></u>

Given that £980 is divided between Caroline, Sarah & Gavyn

Let "c" be the amount received by caroline

Let "s" be the amount received by sarah

Let "g" be the amount received by gavyn

<em><u>Caroline gets twice as much as Sarah</u></em>

amount received by caroline = twice as much as Sarah

amount received by caroline = 2(amount received by sarah)

c = 2s ---- eqn 1

<em><u>Sarah gets three times as much as Gavyn</u></em>

amount received by sarah = three times as much as Gavyn

amount received by sarah = 3(amount received by gavyn)

s = 3g ------- eqn 2

Given that total amount is 980

c + s + g = 980 --- eqn 3

<em><u>Let us solve eqn 1, 2, 3 to get values of "c" "s" "g"</u></em>

From eqn 2,

g = \frac{s}{3}  --- eqn 4

Substitute eqn 1 and eqn 4 in eqn 3

2s + s + \frac{s}{3} = 980\\\\\frac{6s + 3s + s}{3} = 980\\\\6s + 3s + s = 980 \times 3\\\\10s = 2940\\\\s = 294

Thus sarah has received £ 294

7 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
Find the value of x in the triangle.<br> to<br> to
gizmo_the_mogwai [7]

Answer:

45 and 45

Step-by-step explanation:

4 0
3 years ago
Is the square root of 1,815 rational?
Masja [62]

Answer:

It is not

Step-by-step explanation:

4 0
3 years ago
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