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3 years ago

Find the indicated limit, if it exists.

Mathematics

2 answers:

Inessa05 [86]3 years ago

5 0

No solution because it makes the most sense

timama [110]3 years ago

4 0

Answer:

The limit exists

Step-by-step explanation:

First note that for a function to exist, it must be continuous at the point given.

A function therefore said to be continuous if the right hand limit is equal to the left hand limit and equal to the limit of the function at the point x = x0 where x0 is the value of x which the limit is tending towards.

According to the question, the right hand limit function is the corresponding function at when x>-10. The corresponding function is f(x) = x+16

Substituting x = -10 into the function we have;

f(-10) = -10+16

f(-10) = 6

This shows that the right hand limit is 6.

The left hand limit function is the corresponding function at when x<-10. The function is f(x) = -4-x

Substituting x = -10 in the function, we will have;

f(-10) = -4-(-10)

f(-10) = -4+10

f(-10) = 6

This shows that the left hand limit is also 6.

As it can be seen that the corresponding value of the limit at x = -10 is also 6.

Based on the conclusion, since the right hand limit = left hand limit = limit of the function at the point, therefore, the limit of the function is continuous and since since all continuous functions exists, the function above exists

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How to solve this equation of the missing sides of the triangle

iris [78.8K]

Step-by-step explanation:

Given is a 45°-45°-90 ° angled triangle.

By 45°-45°-90 ° angled triangle theorem:

side opposite to 45° angle is $\frac{1}{\sqrt 2}$ times of hypotenuse.

$\therefore \: 7 = \frac{1}{ \sqrt{2} } \times m \\ \\ \huge \red{ \boxed{\therefore \: m \: = 7 \sqrt{2} \: units}} \\ \\ n \: = \frac{1}{ \sqrt{2} } \times m \\ \\ \therefore \:n \: = \frac{1}{ \sqrt{2} } \times 7 \sqrt{2} \\ \huge \purple{ \boxed{\therefore \:n \: = 7 \: units}} \\$

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4 years ago

How to solve this? I don't get it

Neporo4naja [7]

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8 0

3 years ago

The sum of the series {1(2/3)}²+{2(1)3)}²+3²+{3(2/3)}²+....to 10 term is

ryzh [129]

Step-by-step explanation:

<h3>Given Question </h3>

The sum of the series is

$\tt{ {\bigg[1\dfrac{2}{3} \bigg]}^{2} + {\bigg[2\dfrac{1}{3} \bigg]}^{2} + {3}^{2} + {\bigg[3\dfrac{2}{3} \bigg]}^{2} + - - 10 \: terms}$

$\green{\begin{gathered}\large{\sf{{\underline{Formula \: Used - }}}} \end{gathered}}$

$\boxed{\tt{ \displaystyle\sum_{k=1}^{n}1 = n \: }}$

$\boxed{\tt{ \displaystyle\sum_{k=1}^{n}k = \frac{n(n + 1)}{2} \: }}$

$\boxed{\tt{ \displaystyle\sum_{k=1}^{n} {k}^{2} = \frac{n(n + 1)(2n + 1)}{6} \: }}$

$\large\underline{\sf{Solution-}}$

Given series is

$\rm :\longmapsto\: {\bigg[1\dfrac{2}{3} \bigg]}^{2} + {\bigg[2\dfrac{1}{3} \bigg]}^{2} + {3}^{2} + {\bigg[3\dfrac{2}{3} \bigg]}^{2} + - - - 10 \: terms$

can be rewritten as

$\rm \: = \: {\bigg[\dfrac{5}{3} \bigg]}^{2} + {\bigg[\dfrac{7}{3} \bigg]}^{2} + {\bigg[\dfrac{9}{3} \bigg]}^{2} + {\bigg[\dfrac{11}{3} \bigg]}^{2} + - - - 10 \: terms$

$\rm \: = \: \dfrac{1}{9}[ {5}^{2} + {7}^{2} + {9}^{2} + - - - 10 \: terms \: ]$

Now, here, 5, 7, 9 forms an AP series with first term 5 and common difference 2.

So, its general term is given by 5 + ( n - 1 )2 = 5 + 2n - 2 = 2n + 3

So, above series can be represented as

$\rm \: = \: \dfrac{1}{9}\displaystyle\sum_{n=1}^{10}(2n + 3) ^{2}$

$\rm \: = \: \dfrac{1}{9}\displaystyle\sum_{n=1}^{10}\bigg[ {4n}^{2} + 9 + 12n\bigg]$

$\rm \: = \: \dfrac{1}{9}\bigg[\displaystyle\sum_{n=1}^{10} {4n}^{2} + \displaystyle\sum_{n=1}^{10}9 + 12\displaystyle\sum_{n=1}^{10}n\bigg]$

$\rm \: = \: \dfrac{1}{9}\bigg[4\displaystyle\sum_{n=1}^{10} {n}^{2} +9 \displaystyle\sum_{n=1}^{10}1 + 12\displaystyle\sum_{n=1}^{10}n\bigg]$

$\rm \: = \: \dfrac{4}{9}\bigg[\dfrac{10(10 + 1)(20 + 1)}{6} \bigg] + 10 + \dfrac{4}{3}\bigg[\dfrac{10(10 + 1)}{2} \bigg]$

$\rm \: = \: \dfrac{4}{9}\bigg[\dfrac{10(11)(21)}{6} \bigg] + 10 + \dfrac{4}{3}\bigg[\dfrac{10(11)}{2} \bigg]$

$\rm \: = \: \dfrac{1540}{9} + 10 + \dfrac{220}{3}$

$\rm \: = \: \dfrac{1540 + 90 + 660}{9}$

$\rm \: = \: \dfrac{2290}{9}$

Hence,

$\boxed{\tt{ {\bigg[1\dfrac{2}{3} \bigg]}^{2} + {\bigg[2\dfrac{1}{3} \bigg]}^{2} + {3}^{2} + {\bigg[3\dfrac{2}{3} \bigg]}^{2} + - - 10 \: terms = \frac{2290}{9}}}$

6 0

3 years ago

What two numbers add to 9 and multiply to -70

Alborosie

X, y - the numbers

The numbers add to 9 and multiply to -70.
$x+y=9 \\
xy=-70 \\ \\
y=9-x \\
xy=-70 \\ \\
\hbox{substitute 9-x for y in the second equation:} \\
x(9-x)=-70 \\
9x-x^2=-70 \\
-x^2+9x+70=0 \\
-x^2+14x-5x+70=0 \\
-x(x-14)-5(x-14)=0 \\
(-x-5)(x-14)=0 \\
-x-5=0 \ \lor \ x-14=0 \\
x=-5 \ \lor \ x=14 \\ \\
y=9-x \\
y=9-(-5) \ \lor \ y=9-14 \\
y=14 \ \lor \ y=-5 \\ \\
(x,y)=(-5,14) \hbox{ or } (x,y)=(14,-5)$

The numbers are -5 and 14.

4 0

3 years ago

Int var1 = 0b0001; int var2 = 0b1111; int results1 = var1 & var2; int results2 = var1 | var2; int results3 = var1 ^ var2; in

Afina-wow [57]

Int var1 = 0b0001;
int var2 = 0b1111;
int results1 = var1 & var2;
int results2 = var1 | var2;
int results3 = var1 ^ var2;
int printit = results1 + results2 + results3;

what are the values for results1, results2, results3 and printit after executing the code?
notes:
1. faster responses will be obtained if your code is presented line by line (in a file) before posting.
2. please specify language, many languages use the same syntax but could have differences in interpretation
-------------------------------------------------------------------------------

Assuming Java as the language. C is similar.
& bitwise AND &
^ bitwise exclusive OR
| bitwise inclusive OR

So
results1=var1&var2=0b0001&0b1111=0b0001
results2=var1|var2=0b0001&0b1111=0b1111
results3=var1^var2=0b0001&0b1111=0b1110
printit=results1+results2+results3=0b0001+0b1111+0b1110
=0b10000+0b1110
=0b11110

Note: by default, int has 4 signed bytes, ranging from decimal -2147483648 to +2147483647

8 0

4 years ago