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3 years ago

Problem Solving

Mathematics

1 answer:

Murrr4er [49]3 years ago

6 0

Answer:

31 games

Step-by-step explanation:

round 1 has 32 teams ÷2=16games

round 2 has 16 teams ÷2=8games

round 3 has 8 teams ÷2=4games

round 4 has 4 teams ÷ 2=2games

round 5 has 2 teams ÷2=1games

16+8+4+2+1

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A bag of marbles contains 16 blue, 8 green, 9 red, 12 yellow and 5 black marbles. What is the probability of drawing a green mar

Novay_Z [31]

Answer:

Step-by-step explanation:

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3 years ago

Write the quadratic equation whose roots are 5 and 1, and whose leading coefficient is 5

zepelin [54]

Answer: the equation is

5x^2 -30x - 25

Step-by-step explanation:

A quadratic equation is one in which the highest power of the unknown is 2.

The general form of a quadratic equation is expressed as

ax^2 + bx + c

Where c is a constant and a is the leading coefficient

Assuming we want to write the quadratic equation in x, from the information given, the given roots are 5 and 1 and the leading coefficient is 5. We will just multiply the expression by the leading coefficient.

Therefore, the linear factors of the quadratic will be (x-5) and (x-1)

With the leading coefficient as 5, the equation becomes

5(x-5)(x-1)

= 5(x^2 - x - 5x + 5)

= 5(x^2 - 6x + 5)

= 5x^2 -30x - 25

3 0

3 years ago

A rock climber descended 50 feet to a new elevation of 32 feet. What was her initial elevation, e?

alina1380 [7]

Answer:

82 feet

Step-by-step explanation:

50 feet + 32 feet = 82 feet

6 0

3 years ago

Find the product of (3.7 • 10^4) and 2.

Serga [27]

Answer:

7.4 · 10^4

Step-by-step explanation:

For scientific notation, multiplication is pretty simple.

In this case, you would only multiply the "regular" number by 2.

(3.7 · 2) · 10^4

= 7.4 · 10^4

However, if it were asking about multiplying the number with something else in scientific notation, in addition to what we did above, you'd add together the exponents.

I hope this helped!

6 0

3 years ago

In quadratic drag problem, the deceleration is proportional to the square of velocity

Mars2501 [29]

Part A

Given that $a= \frac{dv}{dt} =-kv^2$

Then,

$\int dv= -kv^2\int dt \\ \\ \Rightarrow v(t)=-kv^2t+c$

For $v(0)=v_0$ , then

$v(0)=-kv^2(0)+c=v_0 \\ \\ \Rightarrow c=v_0$

Thus, $v(t)=-kv(t)^2t+v_0$

For $v(t)= \frac{1}{2} v_0$ , we have

$\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\ \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\ \\ \Rightarrow kv_0t=2 \\ \\ \Rightarrow t= \frac{2}{kv_0}$

Part B

Recall that from part A,

$v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\ \\ \Rightarrow dx=-kv^2tdt+v_0dt \\ \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\ \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a$

Now, at initial position, t = 0 and $v=v_0$ , thus we have

$x=a$

and when the velocity drops to half its value, $v= \frac{1}{2} v_0$ and $t= \frac{2}{kv_0}$

Thus,

$x=- \frac{1}{2} k\left( \frac{1}{2} v_0\right)^2\left( \frac{2}{kv_0} \right)^2+v_0\left( \frac{2}{kv_0} \right)+a \\ \\ =- \frac{1}{2k} + \frac{2}{k} +a$

Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

$- \frac{1}{2k} + \frac{2}{k} +a-a \\ \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k}$

7 0

3 years ago