All categories

3 years ago

A right triangle is displayed.

Mathematics

1 answer:

Mazyrski [523]3 years ago

5 0

Answer:

You have to use the Pythagorean Theorem to find the other side, which will be your base in the Area formula (A=1/2bh). So the Pythagorean theorem would be 36^2 + b^2 = 60^2. This gives you 48 in for your base/missing side length. Then, you plug it into the area formula. So, A=1/2(48) (36). This gives you 864 in^2 for the area of your triangle.

You might be interested in

What is 2/3+1/2 equal to

Ahat [919]

2/3 + 1/2 is 7/6
Hope this helps!

5 0

3 years ago

Read 2 more answers

Convert 1 cal/(m^2 * sec * °C) into BTU/(ft^2 * hr * °F)

Crazy boy [7]

Answer:

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

Step-by-step explanation:

To find : Convert $1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}$ into $\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

Solution :

We convert units one by one,

$1\text{ m}^2=10.7639\text{ ft}^2$

$1\text{ sec}=\frac{1}{3600}\text{ hour}$

$1\text{ cal}=0.003968\text{ BTU}$

Converting temperature unit,

$^\circ C\times \frac{9}{5}+32=^\circ F$

$1^\circ C\times \frac{9}{5}+32=33.8^\circ F$

So, $1^\circ C=33.8^\circ F$

Substitute all the values in the unit conversion,

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=\frac{0.003968}{10.7639\times \frac{1}{3600}\times 33.8}\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=\frac{0.003968}{0.101061}\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

Therefore, The conversion of unit is $1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

3 0

3 years ago

Given sin x =-4/5 and x is in quadrent 3, what is the value of tan x/2

4vir4ik [10]

Answer:

We can write sin x in terms of tan x/2 using the formula:

⇒ sin x = (2 tan (x/2)) / (1 + tan2(x/2))

Therefore, using the above formula, we can find the values of tan x/2 by putting the value of sin x.

⇒ -4/5 = (2 tan (x/2)) / (1 + tan2(x/2))

Now, if we replace tan (x/2) by y, we get a quadratic equation:

⇒ 0.8y2 + 2y + 0.8 = 0

⇒ 2y2 + 5y + 2 = 0

By using the quadratic formula, we get y = -0.5, -2

Hence, the value of tan (x/2) = -0.5, -2

Now, we have two solutions of tan (x/2).

Now, let's check for the ideal solution using the formula tan x = (2 tan (x/2)) / (1 - tan2(x/2)).

For tan (x/2) = -0.5:

⇒ tan x = 2(-0.5) / 1 - (-0.5)2 = -4/3

It is also given that x lies in the third quadrant. We know that tan is positive in the third quadrant, and here we get tan x = -4/3 which is negative.

Hence, we can say that tan (x/2) = -0.5 is not a correct solution. Hence it is rejected.

Now let's check for tan (x/2) = -2.

⇒ tan x = 2(-2) / 1 - (-2)2 = 4/3

Here, we get tan x = 4/3 which is positive.

Hence, we can say that tan (x/2) = -2 is a correct solution.

5 0

3 years ago

What is the general form of the equation for the given circle centered at O(0, 0)?

eduard

The general form of such an equation is
x^2 + y^2 = r^2 where r = radius

In this case r^2 = 4^2 + 5^2 = 41

So the required equation is x^2 ^ y^2 = 41

B

4 0

3 years ago

Read 2 more answers

Graph y= 3/4x - 1/2

Elenna [48]

I hope this helps! :)

4 0

3 years ago

Read 2 more answers