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sweet [91]
3 years ago
15

sam writes on a white board the positive integers from 1 to 6 inclusive once each. she then writes p additional fives and q seve

ns on the board. The mean of all the numbers on the board is then 5.3. What is the smallest possible value of q?
Mathematics
2 answers:
Vladimir [108]3 years ago
6 0

Answer:

SAME< I NEEDMHELP!

Step-by-step explanation:

ASAPPP

ASHA 777 [7]3 years ago
5 0

Answer:7

Step-by-step explanation:

This question is on the ukmt junior maths challenge which I'm doing now, out of the options given I got 7, because 2+4+6+5+(5×20)+(7×7)= 166, (then dividing that by the number of numbers to get the mean) so 166÷31= 5.354.....

Also the 3 has a recurring symbol above it

The 7 is times seven so uh yeah<3

You might be interested in
Tickets to the museum cost $7.50 each. Students receive a 15% discount on admission
Ugo [173]

Answer:

$ 6.375

Step-by-step explanation:

Cost of ticket = $7.50

Discount = 15%

Discount price  = 15/100 x 7.50 = 1.125

Cost of student ticket = $7.50 - 1.125 = $ 6.375

I hope im right!!

5 0
3 years ago
How do you solve this
OLga [1]
This is example of what you can do 13-5×=3. -5×=3-13. -5×÷-5=-10÷-5=2. you have to group like times to find ×, then divide both sides with the number that has variable at the side
8 0
4 years ago
Daniel and Sofia both solve this system of equations: y = -1.5x + 5.7 y = 0.5x + 0.3 Daniel's solution is (1.1, 2.4). Sofia's so
scoundrel [369]
Y = -1.5x + 5.7
y = 0.5x + 0.3

-1.5x + 5.7 = 0.5x + 0.3
-1.5x - 0.5x = 0.3 - 5.7
-2x = - 5.4
x = -5.4/-2
x = 2.7

y = 0.5x + 0.3
y = 0.5(2.7) + 0.3
y = 1.35 + 0.3
y = 1.65

solution is (2.7, 1.65)

Sophia is more likely to be correct because the x coordinate is between 2 and 3 and the y coordinate is in between 1 and 2



7 0
3 years ago
Read 2 more answers
Find a point on the curve y= x^2 that is closest to the point (18, 0).? Find a point on the curve y= x^2 that is closest to the
Phoenix [80]
You're trying to minimize the distance between the point (18,0) and an arbitrary point on the curve, (x,y)=(x,x^2).

The distance between two such points is given by the function

d(x)=\sqrt{(x-18)^2+(x^2-0)^2}=\sqrt{x^4+x^2-36x+324}

so this is the function whose derivative you should check.

But before you do that, it's helpful to know that d(x) is minimized at the same point as the modified distance function d^*(x)=d^2(x)=x^4+x^2-36x+324.

Differentiating, you have

\frac{\mathrm d}{\mathrm dx}d^*(x)=4x^3+2x-36

Set this to zero and solve for the critical points.

4x^3+2x-36=2(2x^3+x-18)=0

You can use the rational root theorem to find some potential candidates for roots to the cubic. The constant term has factors \pm1,\pm2,\pm3,\pm6,\pm9,\pm18, while the leading coefficient has factors \pm1,\pm2. The only candidates for rational roots are \pm1,\pm\dfrac12,\pm2,\pm3,\pm\dfrac32,\pm6,\pm9,\pm\dfrac92,\pm18. The only one of these that works is 2, so x=2 is a root to the cubic above.

Polynomial division reveals that we can factor the cubic as

2(x-2)(2x^2+4x+9)=0

which has only one real root at x=2. Checking the value of the derivative of d^* to the left and right of this point confirms that a minimum occurs here.

Therefore the closest point on the curve to (18,0) is (2,4).
6 0
3 years ago
Can someone please help me
Temka [501]
The inverse will have you negate x and y. So you'll stick tilde ~ symbols in front of x and y

So we have x ---> y as the initial expression
The inverse would be ~x ---> ~y

This is why the answer is choice D
4 0
3 years ago
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