Why does calculus exist and I think it really shouldn't?

Jeanette can walk 1 km in 11 minutes. At the same rate, how far can she walk in 55 minutes?

gladu [14]

11 minutes = 1 km

11 x 5 = 55 minutes
1 x 5 = 5 km.

She can walk 5 km in 55 minutes.

3 0

4 years ago

A planet has a circular orbit around a star. It is a distance of 81,000,000 km from the centre of the star. It orbits at an aver

slava [35]

$81,000,000 \text{ km} \left(\frac{1 \text{ h}}{46,000 \text{ km}} \right) =\frac{40500}{23} \text{ h} \\ \\ \frac{40500}{23} \text{ h} \left(\frac{1 \text{ days}}{24 \text{ hours}} \right) \approx \boxed{73 \text{ days}}$

5 0

2 years ago

A Web music store offers two versions of a popular song. The size of the standard version is 2.8 megabytes (MB). The size of the

Serga [27]

Answer:

Number of high quality version downloads = 880

Step-by-step explanation:

Let,

x be the number of standard version songs downloaded

y be the number of high quality version songs downloaded

According to given statement,

x+y=1600 Eqn 1

2.8x+4.1y=5624 Eqn 2

Multiplying Eqn 1 by 2.8 to eliminate x

2.8(x+y=1600)

2.8x+2.8y=4480 Eqn 3

Subtracting Eqn 3 from Eqn 2

(2.8x+4.1y)-(2.8x+2.8y)=5624-4480

2.8x+4.1y-2.8x-2.8y=1144

1.3y=1144

Dividing both sides by 1.3

$\frac{1.3y}{1.3}=\frac{1144}{1.3}\\y=880$

Hence,

Number of high quality version downloads = 880

3 0

4 years ago

ms knight and ms whitaker are orgaizing their summer clothes. ms yanowitz has 6 boxes ms. whitacker has 5 boxes. they want to or

natima [27]

If they have 11 boxes and 88 shorts there will be 8 shorts in each box

4 0

4 years ago

Which equation has the solutions x=1+or-square root of 5?

stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

case a) $x^{2}+2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=-4+1$

$x^{2}+2x+1=-3$

Rewrite as perfect squares

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

$(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i$

therefore

case a) is not the solution of the problem

case b) $x^{2}-2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=-4+1$

$x^{2}-2x+1=-3$

Rewrite as perfect squares

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

$(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i$

therefore

case b) is not the solution of the problem

case c) $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=4+1$

$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

case d) $x^{2}-2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=4+1$

$x^{2}-2x+1=5$

Rewrite as perfect squares

$(x-1)^{2}=5$

$(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}$

therefore

case d) is the solution of the problem

therefore

the answer is

$x^{2}-2x-4=0$

5 0

3 years ago

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