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4 years ago

A pitchers arm rotates at a speed of 7 degrees per millisecond

Mathematics

1 answer:

Fantom [35]4 years ago

7 0

Answer:

To solve this problem, all we have to do is to use a conversion factor to convert millisecond into seconds. We know that there are 1000 milliseconds in a second, therefore:

pitcher’s arm rotation = 7 degrees / millisecond * (1000 milliseconds / second)

pitcher’s arm rotation = 7,000 degrees / second

Step-by-step explanation:

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2(5c+ 2) - 2c = 3(2c + 3) + 7

Mademuasel [1]

Answer:

c = 6

Step-by-step explanation:

Step 1: Write equation

2(5c + 2) - 2c = 3(2c + 3) + 7

Step 2: Solve for <em>c</em>

Distribute: 10c + 4 - 2c = 6c + 9 + 7
Combine like terms: 8c + 4 = 6c + 16
Subtract 6c on both sides: 2c + 4 = 16
Subtract 4 on both sides: 2c = 12
Divide both sides by 2: c = 6

8 0

3 years ago

Read 2 more answers

DO NOT BE DISCOURAGED IF YOU SEE AN ANSWER! PLEASE HELP ASAP WILL GIVE BRAINLIEST AND THANKS AND 5 STAR RATING TO THE CORRECT AN

nignag [31]

See the attached picture:

8 0

3 years ago

If the sum of the zereos of the quadratic polynomial is 3x^2-(3k-2)x-(k-6) is equal to the product of the zereos, then find k?

lys-0071 [83]

Answer:

Step-by-step explanation:

So I'm going to use vieta's formula.

Let u and v the zeros of the given quadratic in ax^2+bx+c form.

By vieta's formula:

1) u+v=-b/a

2) uv=c/a

We are also given not by the formula but by this problem:

3) u+v=uv

If we plug 1) and 2) into 3) we get:

-b/a=c/a

Multiply both sides by a:

-b=c

Here we have:

a=3

b=-(3k-2)

c=-(k-6)

So we are solving

-b=c for k:

3k-2=-(k-6)

Distribute:

3k-2=-k+6

Add k on both sides:

4k-2=6

Add 2 on both side:

4k=8

Divide both sides by 4:

k=2

Let's check:

$3x^2-(3k-2)x-(k-6) \text{ with }k=2$ :

$3x^2-(3\cdot 2-2)x-(2-6)$

$3x^2-4x+4$

I'm going to solve $3x^2-4x+4=0$ for x using the quadratic formula:

$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\frac{4\pm \sqrt{(-4)^2-4(3)(4)}}{2(3)}$

$\frac{4\pm \sqrt{16-16(3)}}{6}$

$\frac{4\pm \sqrt{16}\sqrt{1-(3)}}{6}$

$\frac{4\pm 4\sqrt{-2}}{6}$

$\frac{2\pm 2\sqrt{-2}}{3}$

$\frac{2\pm 2i\sqrt{2}}{3}$

Let's see if uv=u+v holds.

$uv=\frac{2+2i\sqrt{2}}{3} \cdot \frac{2-2i\sqrt{2}}{3}$

Keep in mind you are multiplying conjugates:

$uv=\frac{1}{9}(4-4i^2(2))$

$uv=\frac{1}{9}(4+4(2))$

$uv=\frac{12}{9}=\frac{4}{3}$

Let's see what u+v is now:

$u+v=\frac{2+2i\sqrt{2}}{3}+\frac{2-2i\sqrt{2}}{3}$

$u+v=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}$

We have confirmed uv=u+v for k=2.

4 0

3 years ago

You know that 5^2=25. How can you use this fact to evaluate 5^4

PIT_PIT [208]

Answer:

5^4 = 625

Step-by-step explanation:

5^4 is just 25 times 25

7 0

3 years ago

The function f(t) = t2 + 4t − 14 represents a parabola. Part A: Rewrite the function in vertex form by completing the square. Sh

Cloud [144]

Answer:

A) The vertex ( h , k) = ( -2 , -18)

B) The minimum value of the given function = - 18

C) The Axis of the symmetry for f(t) is y -axis

Step-by-step explanation:

Given a parabola f(t) = t² + 4 t − 14

f(t) = t² + 2(2) t + (2)²-4− 14

f(t) = (t +2)² - 18

Let comparing y = (x +2)² -18

(x +2)² = y + 18

(x-h))² = 4 a ( y - k))²

<em>The vertex ( h , k) = ( -2 , -18)</em>

Given a parabola f(t) = t² + 4 t − 14

Differentiating with respective to 't'

f¹(t) = 2 t + 4

f¹(t) = 2 t + 4 = 0

now $t = \frac{-4}{2} = -2$

Again Differentiating with respective to 't'

$f^{ll} (t) = 2 (1) >0$

f(x) has a minimum value at t = -2

Given f(t) = t² + 4 t − 14

f( -2) = 4 + 4(-2) -14 = 4 -8 -14 = -18

The minimum value of the given function = - 18

f(t) = (t +2)² - 18

Let comparing y = (x +2)² -18

(x +2)² = y + 18

(x-h))² = 4 a ( y - k))²

The vertex ( h , k) = ( -2 , -18)

The Axis of the symmetry for f(t) is y -axis

8 0

3 years ago