28lbs is what % of 58

the cricket team and the football team are sharing the field for their practices today. the cricket team meets for practice ever

polet [3.4K]

Answer:

4 days

Step-by-step explanation:

Crickets team = every 2 days

Football team = every 4 days

How much days from now will they have to share the field again?

Find the lowest common multiple of 2 days and 4 days

2 days = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20

4 days = 4, 8, 12, 16, 20

The lowest common multiple of 2 days and 4 days = 4 days

Therefore, if the football team and Crickets team share the field to practice today, they will have to share the field again in the next 4 days

3 0

3 years ago

Khloe goes to a store an buys an item that costs xx dollars. She has a coupon for 15% off, and then a 8% tax is added to the dis

SSSSS [86.1K]

Answer:

0.93x

Step-by-step explanation:

the total amount is:

x(1-15%+8%)=0.93x

so the answer is 0.93x

7 0

2 years ago

4. Find the standard from of the equation of a hyperbola whose foci are (-1,2), (5,2) and its vertices are end points of the dia

Ad libitum [116K]

The standard form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

<h3>How to find the standard equation of a hyperbola</h3>

In this problem we must determine the equation of the hyperbola in its standard form from the coordinates of the foci and a general equation of a circle. Based on the location of the foci, we see that the axis of symmetry of the hyperbola is parallel to the x-axis. Besides, the center of the hyperbola is the midpoint of the line segment with the foci as endpoints:

(h, k) = 0.5 · (- 1, 2) + 0.5 · (5, 2)

(h, k) = (2, 2)

To determine whether it is possible that the vertices are endpoints of the diameter of the circle, we proceed to modify the general equation of the circle into its standard form.

If the vertices of the hyperbola are endpoints of the diameter of the circle, then the center of the circle must be the midpoint of the line segment. By algebra we find that:

x² + y² - 4 · x - 4 · y + 4= 0

(x² - 4 · x + 4) + (y² - 4 · y + 4) = 4

(x - 2)² + (y - 2)² = 2²

The center of the circle is the midpoint of the line segment. Now we proceed to determine the vertices of the hyperbola:

V₁(x, y) = (0, 2), V₂(x, y) = (4, 2)

And the distance from the center to any of the vertices is 2 (semi-major distance, a) and the semi-minor distance is:

b = √(c² - a²)

b = √(3² - 2²)

b = √5

Therefore, the standard form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

To learn more on hyperbolae: brainly.com/question/27799190

#SPJ1

5 0

2 years ago

A ray of light with intensity lo falls vertically on the sur face of the sea and at ten meters deep its intensity is 10/3. Assum

seropon [69]

Answer:

intensity = $\frac{Io}{15}$

intensity = $\frac{Io}{30}$

depth = 333.33 m

Step-by-step explanation:

given data

deep = 10 m

intensity = Io

intensity = Io/3

to find out

intensity of light at 50 m and 100 m and 1/100 of initial intensity remain ?

solution

we know here that intensity is inversely proportional to deep so

intensity = k × $\frac{1}{Deep}$ .................1

here k is constant

so we have given 10 m deep so

$\frac{Io}{3}$ = $\frac{k}{10}$

so k = Io × $\frac{10}{3}$ ................2

so from equation 1 when 100 m deep and 50 m deep

intensity = k × $\frac{1}{Deep}$

intensity = Io × $\frac{10}{3}$ × $\frac{1}{50}$

intensity = $\frac{Io}{15}$

and

intensity = Io × $\frac{10}{3}$ × $\frac{1}{100}$

intensity = $\frac{Io}{30}$

and

at intensity Io/100

intensity = k × $\frac{1}{Deep}$

$\frac{Io}{100}$ = Io × $\frac{10}{3}$ × $\frac{1}{D}$

D = 333.33 m

so depth = 333.33 m

3 0

3 years ago

Simplify the square root of four over the cubed root of four.

Vikki [24]

$\bf ~\hspace{7em}\textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} ~\hspace{10em} a^{-\frac{ n}{ m}} \implies \cfrac{1}{a^{\frac{ n}{ m}}} \implies \cfrac{1}{\sqrt[ m]{a^ n}} \\\\\\ ~\hspace{7em}\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} ~\hspace{4.5em} \cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \cfrac{\sqrt{4}}{\sqrt[3]{4}}\implies \cfrac{\sqrt[2]{4}}{\sqrt[3]{4}}\implies \cfrac{4^{\frac{1}{2}}}{4^{\frac{1}{3}}}\implies 4^{\frac{1}{2}}\cdot 4^{-\frac{1}{3}}\implies 4^{\frac{1}{2}-\frac{1}{3}}\implies 4^{\frac{3-2}{6}} \\\\\\ 4^{\frac{1}{6}}\implies (2^2)^{\frac{1}{6}}\implies 2^{2\cdot \frac{1}{6}}\implies 2\frac{1}{3}\implies \sqrt[3]{2}$

7 0

3 years ago

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28lbs is what % of 58​

28lbs is what % of 58