All categories

2 years ago

A television screen measures 35 cm wide and 26 cm high. What is the diagonal measure of the screen?

Mathematics

2 answers:

Sphinxa [80]2 years ago

4 0

We calculate the dioglonal using the Pythagorean theorem

$\displaystyle\ C^2=A^2+B^2=> C=\sqrt{A^2+B^2} \ then \\\\C=\sqrt{26^2+35^2} =\sqrt{1225+676} =\sqrt{1901} \approx43.6\\\\Answer : the \:diagonal \:\:length is \:\underline{43.6}$

SpyIntel [72]2 years ago

3 0

Answer:

43.6 (approximately)

Step-by-step explanation:

Diagonal measures of the screen is,

√(35²+26²)

= √(1901)

= 43.6 (approximately)

Answered by GAUTHMATH

You might be interested in

Write an equation of the circle graphed below

dezoksy [38]

Answer:

(-1,1)

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

In decimal form, to the nearest tenth, what is the probability that a randomly selected riverside high school student is in twel

Kitty [74]

Answer: 0.4 which is choice C

====================================

Work Shown

Focus on the "riverside high school" row only

A = number of twelfth graders = 200

B = number of students = 500

P = probability of picking a twelfth grader

P = A/B

P = 200/500

P = 2/5

P = 0.4

8 0

2 years ago

Read 2 more answers

Plz help me

Iteru [2.4K]

Answer:

I think the answer is 60 feet.

Step-by-step explanation:

Because his rate is 2 feet per second and the question asks where he'll be after 30 seconds, the answer has to be 60 feet.

I hope this helps!

5 0

2 years ago

9 ( 3x + 5y - 11x + 8 )

harina [27]

Answer: -72x+45y+72

Step-by-step explanation:

7 0

2 years ago

Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

7 0

2 years ago