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2 years ago

What are the common denominator for 4/5and 1/3

Mathematics

2 answers:

Maru [420]2 years ago

7 0

Answer:

15 is the answer hope this helps

alexgriva [62]2 years ago

5 0

Answer:

Step-by-step explanation:

Given the following question:

You can find the common denominator by multiplying the first fraction by the second fractions denominator, the same goes with the second fraction.

$\frac{4}{5} \times3=\frac{12}{15}$
$\frac{1}{3} \times5=\frac{5}{15}$
$=15$

Hope this helps.

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If △NMO ∼ △XYZ, and the area of △NMO is 147 square feet, find the area of △XYZ

sladkih [1.3K]

The area of △XYZ according to the figure is 363ft²

<h3>Area of triangles</h3>

The formula for calculating the area of a triangle is expressed as:

A = 0.5bh

b is the base
h is the height

<h3>Calculate area of each triangles</h3>

For the triangle △NMO

Area of △NMO = 0.5bh

147 = 0.5b(14/12)

147 = 7b/12

7b = 12 * 147

b = 252

b = 252ft

b = 3024in

Get the base of △XYZ

14/3024 = 22/XZ

XZ = 4,752in

Area of △XYZ = 0.5 (4,752)(22)

Area of △XYZ= 52,272in²

Area of △XYZ = 363ft²

Hence the area of △XYZ is 363ft²

Learn more on area of triangles here: brainly.com/question/17335144

6 0

1 year ago

An object is launched from ground level with an initial velocity of 120 meters per second. For how long is the object at or abov

babunello [35]

Answer:

D) 14 seconds

Step-by-step explanation:

First we will plug 500 in for y:

500 = -4.9t² + 120t

We want to set this equal to 0 in order to solve it; to do this, subtract 500 from each side:

500-500 = -4.9t² + 120t - 500

0 = -4.9t²+120t-500

Our values for a, b and c are:

a = -4.9; b = 120; c = -500

We will use the quadratic formula to solve this. This will give us the two times that the object is at exactly 500 meters. The difference between these two times will tell us when the object is at or above 500 meters.

The quadratic formula is:

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

Using our values for a, b and c,

$x=\frac{-120\pm \sqrt{120^2-4(-4.9)(-500)}}{2(-4.9)}\\\\=\frac{-120\pm \sqrt{14400-9800}}{-9.8}\\\\=\frac{-120\pm \sqrt{4600}}{-9.8}\\\\=\frac{-120\pm 67.8233}{-9.8}\\\\=\frac{-120+67.8233}{-9.8}\text{ or }\frac{-120-67.8233}{-9.8}\\\\=\frac{-57.1767}{-9.8}\text{ or }\frac{-187.8233}{-9.8}\\\\=5.3242\text{ or }19.1656$

The two times the object is at exactly 500 meters above the ground are at 5 seconds and 19 seconds. This means the amount of time it is at or above 500 meters is

19-5 = 14 seconds.

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3 years ago

A room has a maximum capacity of 396 people. How many people are allowed in the room at anytime?

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