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2 years ago

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mike and sam are running laps. mike runs a lap in 5 minutes and sam runs a lap in 6 minutes. if mile and sam begin running at th

e same time, after how many minutes will they cross the starting line at the same time

1 answer:

nalin [4]2 years ago

5 0

answer:

they will never cross....

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I learn maths and I don't know how to solve this problem.

Lyrx [107]

Answer:

As we get closer to x=-3, the limit is approaches -1, on both sides of -3 so this is a good limit so c is 8.

Step-by-step explanation:

In order for a limit to exists, the left side limit must exist and right side limit must exist and on top of that, they must be equal to each other.

In this case, this a piece wise function and by definition of a limit that exist, the limit of this piece wise function as x approaches 3 from either side, must be equal.

We let x=-3 for the top equation.

$( - 3) {}^{2} - 10 = - 1$

So we set the second equation equal to -1.

$- ( {x}^{2} ) + c = - 1$

$- ( - 3 {}^{2} ) + c = - 1$

$- 9 + c = - 1$

$c = 8$

So the value of c is 8. We can prove this by graphing as well.

5 0

3 years ago

How is solving −5x = −6.5 different from solving − 1 5 x = −6.5? How are the solutions related? Complete the explanation.

Alexxandr [17]

Answer:

Step-by-step explanation:

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3 0

3 years ago

Read 2 more answers

Solve the equation for principal values of x. Express solutions in degrees.

Hitman42 [59]

Option C:

x = 90°

Solution:

Given equation:

$\sin x=1+\cos ^{2} x$

<u>To find the degree:</u>

$\sin x=1+\cos ^{2} x$

Subtract 1 + cos²x from both sides.

$\sin x-1-\cos ^{2} x=0$

Using the trigonometric identity: $\cos ^{2}(x)=1-\sin ^{2}(x)$

$\sin x-1-\left(1-\sin ^{2}x\right)=0$

$\sin x-1-1+\sin ^{2}x=0$

$\sin x-2+\sin ^{2}x=0$

$\sin ^{2}x+\sin x-2=0$

Let sin x = u

$u^2+u-2=0$

Factor the quadratic equation.

$(u+2)(u-1)=0$

u + 2 = 0, u – 1 = 0

u = –2, u = 1

That is sin x = –2, sin x = 1

sin x can't be smaller than –1 for real solutions. So ignore sin x = –2.

sin x = 1

The value of sin is 1 for 90°.

x = 90°.

Option C is the correct answer.

3 0

3 years ago

Read 2 more answers

2617 to the nearest thousand

Sliva [168]

Answer:

3000

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

From a thin piece of cardboard 30 in. by 30 in., square corners are cut out so that the sides can be folded up to make a box. Wh

Brilliant_brown [7]

Answer:

Dimensions 5in x 10in x 10in will yeild a box with a max. volume of 500 cubic inches

Step-by-step explanation:

Volume = height x length x width

considering 'x' as the length of the square corners that has been cut out from the cardboard and also, that is height of the cardboard box.

square corners are cut out so that the sides can be folded up to make a box, cardboard sides would reduce by 2x

therefore,

V = x (30-2x) (30-2x) ---> eq(1)

V=( 30x - 2x²) (30-2x)

V= 900x- 60x² - 60x² + 4x³

V= 4x³ - 120 x²+ 900x

Taking derivative w.r.t 'x'

dV/dx = 12x² - 240 x +900

dV/dx = 4 (3x² - 60x +225)

For maximum dV/dx, make it equal zero

dV/dx = 0

so, 4 (3x² - 60x +225)=0

3x² - 60x+225=0 (taking 3 common)

x² - 20x + 75 =0

Solving this quadratic equation

x² - 15x -5x + 75 =0

x(x-15) - 5(x-15) =0

Either (x-15)=0

x=15

Or x-5=0

x = 5

if we substitute x=15 in eq(1), volume becomes zero.

therefore, x cannot be 15

When x= 5

eq(1)=>V = 5 (30-2(5)) (30-2(5))

V= 5 (10) (10)

V= 500 cubic inches,

Therefore, Dimensions 5in x 10in x 10in will yeild a box with a max. volume of 500 cubic inches

7 0

4 years ago