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3 years ago

How do you reduce 135/50

Mathematics

1 answer:

Marrrta [24]3 years ago

4 0

Answer:

27 / 10 or 2.7

Step-by-step explanation:

135/50

To reduce it, do as following:

Divide it by 5, we got:

135/5 / 50/5

= 27 / 10 or 2.7

Hope this helped :3

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The length of the hypotenuse of a 30-60-90 triangle is 28 m. Find the length of the side opposite the 30 angle

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The answer is 14 m

<span>In a 30°-60°-90° triangle, the hypotenuse (c) is twice the length of the shorter leg (a) which is the opposite to the 30 angle:
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c = 2a

We have c = 28 m

So: 28 = 2a

a = 28 / 2

a = 14 m

7 0

3 years ago

Read 2 more answers

A rectangular plot of land that contains 1500 square meters will be fenced and divided into two equal portions by an additional

photoshop1234 [79]

Let the length be x and the width be w

The perimeter will be:
2x+3w=1500

thus
3w=(1500-2x)
w=(1500-2x)/3
w=500-2/3x

The area will be:
A=x*w
A=x(500-2/3x)
A=500x-(2/3)x²

The above is a quadratic equation; thus finding the axis of symmetry we will evaluate for the value of x that will give us maximum area.

Axis of symmetry:
x=-b/(2a)
from our equation:
a=(-2/3) and b=500
thus
x=-500/[2(-2/3)]
x=375
the length will be 375 m

The width will be 250 m

4 0

3 years ago

5. Identify the correct trigonometry formula to use to solve for X.

Anna007 [38]

9514 1404 393

Answer:

(a) cos(55°) = 11/x

Step-by-step explanation:

The marked sides are the hypotenuse (x) and the side adjacent to the given angle (11). Then the relevant trig ratio is ...

Cos = Adjacent/Hypotenuse

cos(55°) = 11/x

4 0

2 years ago

What is the length of the hypotenuse of the triangles

gogolik [260]

Answer:

To find the Hypotenuse the Pythagoras theorem is used

h² = a² + b²

h = hypotenuse

a = 15cm

b = 8cm

h = ?

h² = 15² + 8²

h = √225 + 64

h = √289

h = 17cm

The Hypotenuse is 17cm

Hope this helps.

6 0

3 years ago

A conical water tank with vertex down has a radius of 13 feet at the top and is 21 feet high. If water flows into the tank at a

VLD [36.1K]

Answer:

$\frac{dh}{dt}\approx0.08622\text{ ft/min}$

Step-by-step explanation:

We know that the conical water tank has a radius of 13 feet and is 21 feet high.

We also know that water is flowing into the tank at a rate of 30ft³/min. In other words, our derivative of the volume with respect to time t is:

$\frac{dV}{dt}=\frac{30\text{ ft}^3}{\text{min}}$

We want to find how fast the depth of the water is increasing when the water is 17 feet deep. So, we want to find dh/dt.

First, remember that the volume for a cone is given by the formula:

$V=\frac{1}{3}\pi r^2h$

We want to find dh/dt. So, let's take the derivative of both sides with respect to the time t. However, first, let's put the equation in terms of h.

We can see that we have two similar triangles. So, we can write the following proportion:

$\frac{r}{h}=\frac{13}{21}$

Multiply both sides by h:

$r=\frac{13}{21}h$

So, let's substitute this in r:

$V=\frac{1}{3}\pi (\frac{13}{21}h)^2h$

Square:

$V=\frac{1}{3}\pi (\frac{169}{441}h^2)h$

Simplify:

$V=\frac{169}{1323}\pi h^3$

Now, let's take the derivative of both sides with respect to t:

$\frac{d}{dt}[V]=\frac{d}{dt}[\frac{169}{1323}\pi h^3}]$

Simplify:

$\frac{dV}{dt}=\frac{169}{1323}\pi \frac{d}{dt}[h^3}]$

Differentiate implicitly. This yields:

$\frac{dV}{dt}=\frac{169}{1323}\pi (3h^2)\frac{dh}{dt}$

We want to find dh/dt when the water is 17 feet deep. So, let's substitute 17 for h. Also, let's substitute 30 for dV/dt. This yields:

$30=\frac{169}{1323}\pi (3(17)^2)\frac{dh}{dt}$

Evaluate:

$30=\frac{146523}{1323}\pi( \frac{dh}{dt})$

Multiply both sides by 1323:

$39690=146523\pi\frac{dh}{dt}$

Solve for dh/dt:

$\frac{dh}{dt}=\frac{39690}{146523}\pi$

Use a calculator. So:

$\frac{dh}{dt}\approx0.08622\text{ ft/min}$

The water is rising at a rate of approximately 0.086 feet per minute.

And we're done!

Edit: Forgot the picture :)

3 0

3 years ago