Please help me!

This monitor has a diagonal length of 19 inches and an aspect ratio of 5:4. What are the width and height of the monitor? Round

nasty-shy [4]

<em>Rounded to the nearest tenth of an inch the Witdth of the monitor is 14.8 inches while the Height of the monitor is 11.9 inches.</em>

<h2>Explanation:</h2><h2 />

The aspect ratio is defined as:

<em>"The proportional relationship between the width and height of an image"</em>

It is usually expressed in a mathematical language as:

$x:y \\ \\ x: \ Width \\ \\ y: \ Height$

So, it is true that:

$x:y=5:4 \\ \\ or: \\ \\ \frac{x}{y}=\frac{5}{4} \\ \\ \therefore x=\frac{5y}{4}$

In this case, we know that the monitor has a diagonal length (D):

$D=19in$

So, by Pythagorean theorem:

$D^2=361 \\ \\ 361=x^2+y^2 \\ \\ Since \ x=\frac{5y}{4}$

$D^2=361 \\ \\ 361=x^2+y^2 \\ \\ Since \ x=\frac{5y}{4} \\ \\ 361=\left(\frac{5y}{4}\right)^2+y^2 \\ \\ 361=\frac{25y^2}{16}+y^2 \\ \\ \frac{41}{16}y^2=361 \\ \\ y^2=\frac{261\times 16}{41} \\ \\ y=\sqrt{140.87} \\ \\ y=11.869in$

Finding x:

$x=\frac{5(11.869)}{4} \\ \\ x=14.836in$

<em>Finally, rounded to the nearest tenth of an inch the Witdth of the monitor is 14.8 inches while the Height of the monitor is 11.9 inches.</em>

<h2>Learn more:</h2>

Pythagorean Theorem: brainly.com/question/10182890

#LearnWithBrainly

5 0

3 years ago

Find the length of AB

aniked [119]

FIRST QUESTION
Given points are:
A(2, -4) and B(6, 2)
Now, Use the distance formula.
distance formula = $\sqrt{ (x_{2}- x_{1})^{2} + ( y_{2} - y_{1} )^{2} }$

Now, plug the values into the formula, So,
distance = $\sqrt{ (6- 2)^{2} + ( 2 - (-4))^{2} }$

= $\sqrt{ (6- 2)^{2} + ( 2 +4))^{2} }$

= $\sqrt{ (4)^{2} + ( 6))^{2} }$

= $\sqrt{ 16+36}$

= $\sqrt{52}$

= $2 \sqrt{13}$

So, the length of AB is $2 \sqrt{13}$ .

<span>THIRD QUESTION
</span>Two points given are:
A(3, -2) and B(1, 1)
Also given that B is the midpoint of AC.

Let, the co-ordinates of C be C(a, b).
Now, using midpoint formula,
Midpoint = $(\frac{ x_{1}+ x_{2} }{2} , \frac{ y_{1}+ y_{2} }{2} )$

$(1, 1)=(\frac{ 3+ a }{2} , \frac{ -2+b }{2} )$

Now, equaling the ordered pair, we have,

$1=\frac{ 3+ a }{2}$ .............equation (1)

$1=\frac{ -2+b }{2}$ ................equation (2)

Now, taking equation (1)
$1=\frac{ 3+ a }{2}$

$1*2=3+a$

$2-3=a$

$a=-1$

Now, taking equation (2)
$1=\frac{ -2+b }{2}$

$1*2=-2+b$

$2+2=b$

$b=4$

<span>So, the co ordinates of C are (a, b) which is <u>(-1 , 4)

</u></span>
<span>SECOND QUESTION:
</span>Given equations are:
2x + 3y = 14.....................equation (1)
-4x + 2y = 4 .....................equation (2)
Taking equation (2)
-4x + 2y = 4
2y = 4 + 4x
y = (4 + 4x) / 2
y = 2 + 2x .......................equation (3)
Now, Taking equation (1)
2x + 3y = 14
Substituting the value of y from equation (3), we get,
2x + 3(2 + 2x) = 14
2x + 6 + 6x = 14
8x = 14 - 6
x = (14 - 6) / 8
x = 1

Taking equation (3)
y = 2 + 2x
Now, substituting the value of x in equation (3), we get,
y= 2 + 2 (1)
y = 2 + 2
y = 4

So, x=1 and y=4

6 0

3 years ago

!!!WILL MARK THE BRAINIEST!!!

Nina [5.8K]

Answer:

$1/65536$