If f(x) = x, then f(64) is equal to a. 1/4 b. -8 c. -4 d. 4

at a small school, one male and one female are selected to represent the student body at the PTSCA meetings. if there are 41 mal

I am Lyosha [343]

It is given that there are 41 males and 48 females in the small school.

So, the number of ways a male student can be chosen from 41 males is $^{41}C_1$

Likewise, the number of ways a female student can be chosen from 48 females is $^{48}C_1$ .

Thus, the total number of ways in which 2-person combinations are possible to represent the student body at the PTSAC meetings will be given by:

$^{41}C_1\times ^{48}C_1=1968$

7 0

2 years ago

Solve for x, y, and z

zloy xaker [14]

Answer:

The solution is x=1,y=2,z=3

Step-by-step explanation:

The given system of equations is ;\

2x−3y+4z=8...(1)

3x+4y−5z=−4...(2)

4x−5y+6z=12...(3)

Make x the subject in equation (1)

$x=\frac{8+3y-4z}{2}...(4)$

Put equation (4) into equation (2) and (3)

$3(\frac{8+3y-4z}{2})+4y-5z=-4$

Multiply through by;

$3(8+3y-4z)+8y-10z=-8$

Expand;

$24+9y-12z+8y-10z=-8$

Simplify;

$17y-22z=-32...(5)$

Equation (4) in (3)

$4(\frac{8+3y-4z}{2})-5y+6z=12$

$2(8+3y-4z)-5y+6z=12$

$16+6y-8z-5y+6z=12$

$y-2z=-4$

$y=2z-4...(6)$

Put equation (6) into equation (5)

$17(2z-4)-22z=-32$

$34z-68-22z=-32$

$34z-22z=-32+68$

$12z=36$

z=3

Put z=3 into equation (6)

y=2(3)-4=2

Put y=2 and z=3 into equation 4

$x=\frac{8+3(2)-4(3)}{2}=1$

The solution is x=1,y=2,z=3

3 0

2 years ago

Find the solutions of the quadratic equation <img src="https://tex.z-dn.net/?f=-x%5E2%2B7x-14%3D0" id="TexFormula1" title="-x^2+

Vinvika [58]

Answer:

Step-by-step explanation:

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3 0

2 years ago

Short on time, please help!

kvv77 [185]

56.3% hope it helped ok

3 0

2 years ago

How can I use math to explain why other people's work is right or wrong?

valentinak56 [21]

Answer:

you can use math to explain others wrong by maybe doing calculations

Step-by-step explanation:

6 0

2 years ago