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3 years ago

If ABF= (7x + 20), FBC=(2x - 5), and ABC= 159", find the value of x. x =

Mathematics

1 answer:

s2008m [1.1K]3 years ago

3 0

Given :

∠ABF = ( 7x + 20 )°

∠FBC=(2x - 5)°

∠ABC= 159°

To Find :

The value of x .

Solution :

Assuming B the center of a circle .

So , all three given angles must be added to 360° .

$(7x+20)+(2x-5)+159=360\\\\9x+15+159=360\\\\x=20.67$

Therefore , the value of x is 20.67° .

Hence , this is the required solution .

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3 years ago

The equation is________<br> Solve the equation x=________

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Answer:

7x-5 = 9

x = 2

Step-by-step explanation:

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3 years ago

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

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3 years ago

What is the value of g(9)?

ryzh [129]

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3 years ago

Domain and Range for the function f(x)=5IXI is

shutvik [7]

Answer:

The domain of the function f(x) is:

$\mathrm{Domain\:of\:}\:5\left|x\right|\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:$

The range of the function f(x) is:

$\mathrm{Range\:of\:}5\left|x\right|:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}$

Step-by-step explanation:

Given the function

$f\left(x\right)=5\left|x\right|$

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We know that the domain of the function is the set of input or arguments for which the function is real and defined.

In other words,

Domain refers to all the possible sets of input values on the x-axis.

It is clear that the function has undefined points nor domain constraints.

Thus, the domain of the function f(x) is:

$\mathrm{Domain\:of\:}\:5\left|x\right|\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:$

Determining the range:

We also know that range is the set of values of the dependent variable for which a function is defined.

In other words,

Range refers to all the possible sets of output values on the y-axis.

We know that the range of an Absolute function is of the form

$c|ax+b|+k\:\mathrm{is}\:\:f\left(x\right)\ge \:k$

$k=0$

Thus, the range of the function f(x) is:

$\mathrm{Range\:of\:}5\left|x\right|:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}$

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3 years ago