What is the surface area of a sphere with a radius of 19 units?

Answer:

48 inches

Step-by-step explanation:

The area of a square is given by

A = s^2 where s is the side length

144 = s^2

Taking the square root of each side

sqrt(144) = sqrt(s^2)

12 = s

The side length is 12

We want to find the perimeter to determine the amount of ribbon

P =4s

P =4*12

48 inches

Answer:18

Step-by-step explanation:

Answer:

D

Step-by-step explanation:

Replace R1 with 1 and R3 with 0 you have -2(1)+0 which equals 0 and that matches matrix II. Repeat the process with the other numbers in R1 and R3 and they all come out equal therefor the answer is D.

Answer:

$18,720.5

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Depreciation Formula: A = P(1 - rt)

Step-by-step explanation:

Step 1: Define

Identify variables

P = 36001

r = 16% = 0.16

t = 3

Step 2: Find Cost

Answer:

The second choice is correct

Step-by-step explanation:

Go to a point like 4,3 and 0,1.   the difference in x is 4 since 4-0 = 4.

4,3 and 0,1 now we find the change in y, 3-1 which is 2. the formula for slope is change in y over change in x so 2/4 is the slope or 1/2.

The y intercept of the line is 1 since the line touches the y-axis at the y, 1.

therefore the equation is y = 1/2x + 1.

Answer:

(0,-41)

Step-by-step explanation:

Let see if this table of values are proportional.

This mean that the function is proportional. This also gives us the slope which is 19/9.

This tell us that this function will be a linear function since we are trying to find the y intercept we would use this equation

We know the slope and we know a point so let use point slope form to find the intercept.

[tex]y - y _1= m(x - x _1)[/tex]

Let use points (-18,-79) and let m =19/9.

[tex]y + 79 = \frac{19}{9} (x + 18)[/tex]

[tex]y + 79 = \frac{19}{9} x + 38[/tex]

[tex]y = \frac{19}{9} x - 41[/tex]

This is in slope intercept form so the y intercept is

[tex] - 41[/tex]

The answer is -41.

Answer

16.25

Step-by-step explanation: To The Nearest Hundredth

(-1, 2), (3, 1)

√17

(3, 1), (7, 2)

√17

(3, 1), (7, 2)

√8

So our answer is =

8 + √17 + √17 = 16.25

hope it works for you!

Answer

supplementary angles sum to 180

1 & 2

1 & 4

1 & 8

1 & 6

2 & 3

2 & 7

2 & 5

3 & 4

3 & 8

3 & 6

4 & 5

4 & 7

5 & 6

5 & 8

6 & 7

8 & 7

Answer:

9

Step-by-step explanation:

a^2 +b

Let a = -2 and b = 5

(-2)^2 +5

Exponents first

4 +5

Then add

9

Answer:

9

Step-by-step explanation:

Here, we have to observe that a^2 = a×a =-2 × -2 = 4

hence, a^2 + b = 4 + 5 = 9.

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

Explanation:

Pick any two rows from the table to plug into the slope formula.

I'll pick the rows where every value is positive (rows 3 and 4)

Using the slope formula, we get the following:

m = (y2-y1)/(x2-x1)

m = (1-5)/(2-1)

m = -4/1

m = -4 is the slope and it's the rate of change

For any linear function, the slope and rate of change are the same thing.

Answer:

6 a minus 4 b + 11

Step-by-step explanation:

(8a -3b +7) - (2a + b -4)

= 8a -3b +7 -2a-b +4

= 6a -4b + 11

Answer:

YEAH!!!!!! ITS 45

Step-by-step explanation:

MHM!

Answer:

woooooooooooooooohoooooooooo points go brrrrrr

Step-by-step explanation:

!!

Answer:

It is not valid since there is more than one variable being changed at one.

Step-by-step explanation:

Jackson changed the brand of soil, but he also changes the amount of water, and the amount of sunlight the plants are receiving, so his conclusion that the soil was more effective is not valid.

Answer:

No, Jackson did not take into account alternate theories.

Jackson's findings could have been influenced by the watering schedule and the amount of sunlight. Without adjusting for these variables, the experiment is inconclusive.

Answer:

X = 3.6

Step-by-step explanation:

tan x = O/A

tan 20° = x/10

x = tan 20°(10)

x = 3.6

Answer:

x = 3.6 cm

Step-by-step explanation:

Using trigonometric ratios :

        [tex]sin \theta = \frac{ opposite} {hypotenuse} \\\\cos \theta = \frac{adjacent }{hypotenuse}\\\\tan \theta = \frac{opposite}{adjacent}\\\\[/tex]

[tex]where \ \theta = 20, \ opposite = x \ , hypotenuse = y \ , \ adjacent = 10cm[/tex]

[tex]we \ will \ take \ tan \ 20 , \ because\ we \ have \ to \ find \ opposite \ and \ adjacent \ is\ given.\\\\tan 20 = \frac{x}{10}\\\\x = 10 \times tan 20\\\\x = 10 \times 0.36397\\\\x = 3.6397\\\\x = 3.6 \ cm[/tex]

Answer:

El área del gimnasio sería 2000 metros cuadrados

Step-by-step explanation:

Un lado del gimnasio será igual al lado de la cancha y el otro lado será igual al lado de la piscina.

Dado que tanto la cancha como la piscina tienen forma cuadrada. El lado de la cancha será raíz cuadrada de 2.500 m2 y el de la piscina será raíz cuadrada de 1.600 m2

Por lo tanto, el lado del gimnasio sería 50 y 40.

El área del gimnasio sería 50 * 40 = 2000 metros cuadrados

Answer:

The chances of her pulling the candy out in that exact order for all of them is extremley low

Step-by-step explanation:

I cannot figure out D though sorry

Answer:

Your welcome

Step-by-step explanation:

Answer:

14cm

Step-by-step explanation:

The shape is a rectangle.

The formula for perimeter of a rectangle= 2(lenght + breadth)

lenght = 4cm

width = 3cm

2(4+3)

2 × 7

= 14cm

Answer:

B

Step-by-step explanation:

The triangles do have equal angles and proportional sides, because they are bot right triangles formed on parallel lines.

:)ur welcome

Answer: Choice B

Yes. Right triangles formed on parallel lines have proportional side lengths and congruent angles.

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

Explanation:

Parallel lines always have equal slopes, but different y intercepts.

Recall that slope = rise/run.

The "rise" is the vertical portion of the triangle, while the "run" is the horizontal portion.

So let's say we had a slope of rise/run = 2/3. This means rise = 2 and run = 3.

If we had something like rise/run = 4/6, then that reduces to 2/3, showing that 4/6 = 2/3. They are the same slope.

We can use the SAS similarity theorem to prove the triangles formed in the diagram are congruent based on what is discussed above.

Because the triangles are similar, this means the corresponding angles are congruent. One such pair is the pair of right angles shown.

Answer:

x= -7

Step-by-step explanation:

the labeled angles are alternate-interior angles which are congruent

we can solve for 'x' by creating this equation:

x+139 = 132

x = -7

Answer:

177 kg

Step-by-step explanation:

Find the weight of the third article by finding the difference between 502 and 856, since 856 kg is the total weight and 502 kg is the weight of the other 2 articles:

856 - 502

= 354

Now, find 1/2 of the weight, by dividing this by 2:

354/2

= 177

So, 1/2 the weight of the third article is 177 kg

Answer:

A

Step-by-step explanation:

Answer:

A

Step-by-step explanation:

If you just plug in the answers, so lets take (1,6) and plug it into y = 6x

You could just solve it so it'd be like

6 = 6(1) and

12 = 6(2)

Answer:

157

Step-by-step explanation:

Let the number Angad was thinking be 'x'

Equation:-

1/2(x) = 78.5

x = 78.5(2)

x = 157

Answer:

0.36,0.44,7/10

Step-by-step explanation:

Answer:

0.36, 0.44, 7/10

Step-by-step explanation:

7/10=0.7=0.70

B

Please mark me as a brainlist

Answer:

900

Step-by-step explanation:

3²×5 = 45

3×2²×5² = 300

LCM = 900

Hope this helps... pls vote as brainliest

Answer:

volumn= l*b*h

4x*2x*1x= 8000 cm3

8x= 8000 cm3

x= 1000cm3

length = 4000

breadth= 2000

height = 1000

Answer:

[tex] \displaystyle - \frac{1}{2} [/tex]

Step-by-step explanation:

we would like to compute the following limit:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{1}{ \ln(x + \sqrt{ {x}^{2} + 1} ) } - \frac{1}{ \ln(x + 1) } \right) [/tex]

if we substitute 0 directly we would end up with:

[tex] \displaystyle\frac{1}{0} - \frac{1}{0} [/tex]

which is an indeterminate form! therefore we need an alternate way to compute the limit to do so simplify the expression and that yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \ln(x + 1) - \ln(x + \sqrt{ {x}^{2} + 1 } }{ \ln(x + \sqrt{ {x}^{2} + 1} ) \ln(x + 1) } \right) [/tex]

now notice that after simplifying we ended up with a rational expression in that case to compute the limit we can consider using L'hopital rule which states that

[tex] \rm \displaystyle \lim _{x \to c} \left( \frac{f(x)}{g(x)} \right) = \lim _{x \to c} \left( \frac{f'(x)}{g'(x)} \right) [/tex]

thus apply L'hopital rule which yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \dfrac{d}{dx} \ln(x + 1) - \ln(x + \sqrt{ {x}^{2} + 1 } }{ \dfrac{d}{dx} \ln(x + \sqrt{ {x}^{2} + 1} ) \ln(x + 1) } \right) [/tex]

use difference and Product derivation rule to differentiate the numerator and the denominator respectively which yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \frac{1}{x + 1} - \frac{1}{ \sqrt{x + 1} } }{ \frac{ \ln(x + 1)}{ \sqrt{ {x}^{2} + 1 } } + \frac{ \ln(x + \sqrt{x ^{2} + 1 } }{x + 1} } \right) [/tex]

simplify which yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \sqrt{ {x}^{2} + 1 } - x - 1 }{ (x + 1)\ln(x + 1 ) + \sqrt{ {x}^{2} + 1} \ln( x + \sqrt{ {x }^{2} + 1} ) } \right) [/tex]

unfortunately! it's still an indeterminate form if we substitute 0 for x therefore apply L'hopital rule once again which yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \dfrac{d}{dx} \sqrt{ {x}^{2} + 1 } - x - 1 }{ \dfrac{d}{dx} (x + 1)\ln(x + 1 ) + \sqrt{ {x}^{2} + 1} \ln( x + \sqrt{ {x }^{2} + 1} ) } \right) [/tex]

use difference and sum derivation rule to differentiate the numerator and the denominator respectively and that is yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \frac{x}{ \sqrt{ {x}^{2} + 1 } } - 1}{ \ln(x + 1) + 2 + \frac{x \ln(x + \sqrt{ {x}^{2} + 1 } ) }{ \sqrt{ {x}^{2} + 1 } } } \right) [/tex]

thank god! now it's not an indeterminate form if we substitute 0 for x thus do so which yields:

[tex] \displaystyle \frac{ \frac{0}{ \sqrt{ {0}^{2} + 1 } } - 1}{ \ln(0 + 1) + 2 + \frac{0 \ln(0 + \sqrt{ {0}^{2} + 1 } ) }{ \sqrt{ {0}^{2} + 1 } } } [/tex]

simplify which yields:

[tex] \displaystyle - \frac{1}{2} [/tex]

finally, we are done!

9514 1404 393

Answer:

  -1/2

Step-by-step explanation:

Evaluating the expression directly at x=0 gives ...

  [tex]\dfrac{1}{\ln(\sqrt{1})}-\dfrac{1}{\ln(1)}=\dfrac{1}{0}-\dfrac{1}{0}\qquad\text{an indeterminate form}[/tex]

Using the linear approximations of the log and root functions, we can put this in a form that can be evaluated at x=0.

The approximations of interest are ...

  [tex]\ln(x+1)\approx x\quad\text{for x near 0}\\\\\sqrt{x+1}\approx \dfrac{x}{2}+1\quad\text{for x near 0}[/tex]

__

Then as x nears zero, the limit we seek is reasonably approximated by the limit ...

  [tex]\displaystyle\lim_{x\to0}\left(\dfrac{1}{x+\dfrac{x^2}{2}}-\dfrac{1}{x}\right)=\lim_{x\to0}\left(\dfrac{x-(x+\dfrac{x^2}{2})}{x(x+\dfrac{x^2}{2})}\right)\\\\=\lim_{x\to0}\dfrac{-\dfrac{x^2}{2}}{x^2(1+\dfrac{x}{2})}=\lim_{x\to0}\dfrac{-1}{2+x}=\boxed{-\dfrac{1}{2}}[/tex]

_____

I find a graphing calculator can often give a good clue as to the limit of a function.

Answer:

The domain is all real number

Answer:1/6

Step-by-step explanation:

[tex]\huge\textsf{Hey there!}[/tex]

[tex]\mathsf{\dfrac{4}{5}=\boxed{\bf 0.80\ or\ 80\%}}\\\\\\\mathsf{\dfrac{1}{6}=\boxed{\bf 0.166667\ or\ 16.6667\% }}\\\\\\\mathsf{\dfrac{2}{3}=\boxed{\bf 0.666667\ or \ 66.6667\%}}\\\\\\\mathsf{\dfrac{1}{5}=\boxed{\bf 0.20\ or \ 20\%}}[/tex]

[tex]\large\textsf{After converting the fractions to decimals and percentages, we can}\\\large\textsf{now out them from descending (least) to ascending (greatest) order}\\\large\textsf{to determine your result to this question}[/tex]

[tex]\boxed{\mathsf{\dfrac{1}{6}, \ \dfrac{1}{5}, \ \dfrac{2}{3}, \& \ \dfrac{4}{5}}}[/tex]

[tex]\large\textsf{The SMALLEST one out of your set is: }\boxed{\mathsf{\bf \dfrac{1}{6}}}\large\checkmark[/tex]

[tex]\boxed{\boxed{\large\textsf{Answer: }\mathsf{\bf \dfrac{1}{6}}}}\huge\checkmark[/tex]

[tex]\large\textsf{Good luck on your assignment and enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]