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3 years ago

Someone please helpppo meeeeee !!! ASAP

Mathematics

1 answer:

Wewaii [24]3 years ago

4 0

Because the left-hand side is always >0, the equation has No Solution.

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How to find r in this equation using combination formula C(8,r)=28

slavikrds [6]

Answer:

r = 2

Step-by-step explanation:

We have the formula of $^nC_r = \frac{n!}{r! (n-r)!}$

Now, it is given that $^8C_r = \frac{8!}{r! (8-r)!} = 28$ ........ (1)

And we have to find the value of r which satisfy the above equation.

So, $r! (8-r)! = \frac{8!}{28} = \frac{40320}{28} = 1440$

Now, we have to use the trial method to find the value of r.

For r = 1, $1! (8-1)! = 7! = 5040 \neq 1440$

Hence, r can not be 1.

Now, put r = 2, $2! (8-2)! = 2 \times 6! = 1440$

Therefore, r = 2 (Answer)

5 0

3 years ago

Read 2 more answers

Define the function y(t) to be x(t) delayed by 0.3 seconds. Plot both x(t) and y(t) on the same graph using different line style

Andreas93 [3]

Answer:

$y(t) =x(t+0.3s)$

Step-by-step explanation:

Define the function y(t) to be x(t) delayed by 0.3 seconds.

To interpret that we can say that y at time t take the value of x at time (t+0.3s).

So we can write

$y(t) =x(t+0.3s)$

The graph relates X(t) and Y(t) for t=0.3 to t=4.0. There is a linear relation between the X and Y because the slope Y/X is a constant.

7 0

3 years ago

Keep Trying Write two division problems that have zeros in the quotient. One of the problems should have a remainder and the oth

atroni [7]

So basically you must first use the step that got you here

7 0

3 years ago

HELPPPPP! WILL Give BRAINLIEST! 100 POINTS! Please get it correct!

Leno4ka [110]

Answer:

11.26 cm²

Step-by-step explanation:

The shaded region is the difference of areas of semicircle AB and triangle ABC.

ABC is right triangle as AB is diameter, so C is right angle.

Area of semicircle:

A = 1/2πr² = 1/2*3.14*4² = 25.12 cm²

Area of triangle:

A = 1/2BC*AC = 1/2*4* $\sqrt{8^2-4^2}$ = 2 $\sqrt{48}$ = 13.86 cm²

Shaded are is:

25.12 - 13.86 = 11.26 cm²

4 0

3 years ago

Read 2 more answers

I need help for these problems!!! Thanks!

Triss [41]

I do before but your questions a bit wired seems like they already help u solve ?

8 0

4 years ago