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2 years ago

Help with those 3 please.. ty

Mathematics

2 answers:

irinina [24]2 years ago

7 0

I really just needed the points no cap monkey

storchak [24]2 years ago

7 0

For the second question:

Newton's first law states that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.

For the third question:

Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

Sorry, I couldn’t get the first question.

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Explanation and answer please :)

puteri [66]

Step-by-step explanation:

You use the I = PRN formula

P = 11 500

R = 7.25% but you have to 7.25 divide by 100 so it becomes a decimal so it should be 0.0725

N = 2 years and 6 months. This can be 2.5 years because half of 12 is 6

the solution is 11 500 × 0.0725 × 2.5 = 2084.38

7 0

2 years ago

Jermaine is saving $60 per week for a new guitar. The amount he has saved, s, is represented by

german

Answer:

Step-by-step explanation:

8 0

2 years ago

☆pls help no one has helped me and I'm so confused!

motikmotik

Answer:

Step-by-step explanation:

a). $\frac{5\pm\sqrt{-4}}{3}$ = $\frac{5}{3}\pm \frac{\sqrt{-4}}{3}$

= $\frac{5}{3}\pm\frac{2\sqrt{(-1)} }{3}$

= $\frac{5}{3}\pm\frac{2i}{3}$ [Since, i = $\sqrt{(-1)}$ ]

b). $\frac{10\pm\sqrt{-16}}{2}$ = $\frac{10}{2}\pm \frac{\sqrt{-16}}{2}$

= $5\pm \frac{4\sqrt{-1}}{2}$

= 5 ± 2i [Since, i = $\sqrt{(-1)}$ ]

c). $\frac{-3\pm \sqrt{-144}}{6}$ = $-\frac{3}{6}\pm \frac{\sqrt{-144}}{6}$

= $-\frac{3}{6}\pm \frac{12\sqrt{-1}}{6}$

= $-\frac{1}{2}\pm 2\sqrt{-1}$

= $-\frac{1}{2}\pm 2i$ [Since, i = $\sqrt{(-1)}$ ]

4 0

2 years ago

Complete parts a through c for the given function.

victus00 [196]

Answer:

a. The critcal points are at

$x=0,-5,3$

b. Then, $x = -5$ is a maximum and $x=3$ is a minimum

c. The absolute minimum is at $x = 3$ and the absolute maximum is at $x = -5.$

Step-by-step explanation:

(a)

Remember that you need to find the points where

$f'(x)=0$

Therefore you have to solve this equation.

$20x^4 + 40x^3 - 300x^2 = 0$

From that equation you can factor out $20x^2$ and you would get

$20x^2 ( x^2 +2x - 15) = 0$

And from that you would have $20x^2 = 0$ , so $x = 0$ .

And you would also have $x^2 +2x-15 = 0$ .

You can factor that equation as $x^2 +2x -15 = (x+5)(x-3) = 0$

Therefore $x=-5 , x=3$ .

So the critcal points are at

$x=0,-5,3$

Remember that a function has a maximum at a critical point if the second derivative at that point is negative. Since

$f''(x) = 80x^3 + 120x^2 -600x\\f''(-5) = 80(-5)^3 + 120(-5)^2 -600(-5) = -4000 < 0\\\\f''(3) = 80(3)^3 + 120(3)^2 -600(3) = 1440 > 0 \\$

Then, $x = -5$ is a maximum and $x=3$ is a minimum

The absolute minimum is at $x = 3$ and the absolute maximum is at $x = -5.$

6 0

2 years ago

Read 2 more answers

Curtis bought a skateboard which he later sold at a 30% loss for £14. How much did Curtis pay for

Norma-Jean [14]

Answer:

18 pounds if I'm correct

3 0

2 years ago