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3 years ago

What is the spuare root of 144

Mathematics

2 answers:

SVEN [57.7K]3 years ago

6 0

Answer:

The square root of 144 will be 12

sveticcg [70]3 years ago

3 0

Answer:

12 is the answer

Step-by-step explanation:

12x12=144

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Do you know the answer ??

bazaltina [42]

Answer: The answer is 6

7 0

3 years ago

What is the area of this rectangle? Provide supportive evidence so I can learn better about this. I’m having troubles.

Katyanochek1 [597]

Answer:

the answer is 15x^8 y^3

Step-by-step explanation:

the area of a rectangle is length times width right? this means you have to multiply the values that represent the length and width. in this case those values would be 5x^6 y^2 and 3x^2 y

step 1. set up your problem

it should look like this:

(5x^6 y^2)(3x^2 y)

step 2. multiply the values with corresponding variables

first, x values

(5x^6)(3x^2)

5×3=15 and x^6 × x^2=x^8 when multiplying exponents you add the powers the numbers are raised to. in this case 6+2=8

you should end up with 15x^8

next, y values

y^2 + y

remember what i said about multiplying exponents? you just add the power the variables are raised to. in this case, 2 and 1. when their is no exponent we just add a 1. so, 2+1=3

you end up with y^3

lastly, combine

15x^8 + y^3 = 15x^8y^3

4 0

3 years ago

Help please need asap

yaroslaw [1]

Answer:

A. Obtuse B. Right C. Acute D. Acute Hope this helps :)

7 0

3 years ago

The amount of time a certain brand of light bulb lasts is normally distributed with a

scoundrel [369]

Answer:

92.65 % light bulbs last less than 2080 hours

Step-by-step explanation:

We are given that The amount of time a certain brand of light bulb lasts is normally distributed

Mean = $\mu = 2000$

Standard deviation = $\sigma = 55$

We are supposed to find What percentage light bulbs last less than 2080 hours, to the nearest tenth

$Z = \frac{x-\mu}{\sigma}\\\\Z=\frac{2080-2000}{55}$

Z=1.45

Refer the z table for p value

p value = 0.9265

Hence 92.65 % light bulbs last less than 2080 hours

5 0

3 years ago

Find the derivative of <img src="https://tex.z-dn.net/?f=tan%5E%7B-1%7D%20x" id="TexFormula1" title="tan^{-1} x" alt="tan^{-1} x

sladkih [1.3K]

$\huge{\color{magenta}{\fbox{\textsf{\textbf{Answer}}}}}$

$\frak {\huge{ \frac{1}{1 + {x}^{2} } }}$

Step-by-step explanation:

$\sf let \: f(x) = { \tan }^{ - 1} x \\ \\ \sf f(x + h) = { \tan}^{ - 1} (x + h)$

$\sf f'(x) = \frac{f(x+h) - f(x) }{h}$

$\sf \implies \lim_{ h \to 0 } \frac{ { \tan }^{ - 1}(x + h) - { \tan}^{ - 1}x }{h} \\ \\ \\ \sf \implies \lim_ {h \to 0} \frac{ { \tan}^{ - 1} \frac{x + h - x}{1 + (x + h)x} }{h}$

By using

$\sf { \tan}^{ - 1} x - { \tan}^{ - 1} y = \\ \sf { \tan}^{ - 1} \frac{x - y}{1 + xy} formula$

$\sf \implies \large \lim_{h \to0 } \frac{ { \tan}^{ - 1} \frac{h}{1 + hx + {x}^{2} } }{h} \\ \\ \\ \sf \implies \large{\lim_{h \to0} } \frac{ { \tan}^{ - 1} \frac{h}{1 + hx + {x}^{2} } }{ \frac{h}{1 + hx + {x}^{2} } \times (1 + hx + {x}^{2} )} \\ \\ \\ \sf \implies \large \lim_{h \to0} \frac{ { \tan}^{ - 1} \frac{h}{1 + hx + {x}^{2} } }{ \frac{h}{1 + hx + {x}^{2} } } + \lim_{h \to0} \frac{1}{1 + hx + {x}^{2} }$

Now putting the value of h = 0

$\sf \large \implies 0 + \frac{1}{1 + 0 + {x}^{2} } \\ \\ \\ \purple{ \boxed { \implies \frac{1}{1 + {x}^{2} } }}$

6 0

2 years ago

What is the spuare root of 144​

What is the spuare root of 144