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irinina [24]
3 years ago
10

In the figure, ABCDE is a pentagon with BE||CD and BCIDE. BC is perpendicular to

Mathematics
1 answer:
DiKsa [7]3 years ago
5 0

Answer:

BC=DE= x-y

AB+BC+CD+DE=5+x-y+x+y+x-y=27 cm

5+3x-y=27 cm

3x-y=22

if x=9 , then 3*9-y=22,  27-y=22, y=27-22, y=5

Step-by-step explanation:

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If c(x)= <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B5%7D%7Bx-2%7D%20" id="TexFormula1" title=" \frac{5}{x-2} " alt=" \frac{5
ale4655 [162]
\bf \begin{cases}&#10;c(x)=\cfrac{5}{x-2}\\\\&#10;d(x)=x+3\\&#10;----------\\&#10;(cd)(x)\implies c(x)\cdot d(x)&#10;\end{cases}&#10;\\\\\\&#10;c(x)\cdot d(x)\implies \left( \cfrac{5}{x-2}\right)(x+3)\implies \cfrac{5(x+3)}{x-2}

now, for a fraction, if its denominator ever becomes 0, the fraction becomes undefined, because it'd be a division by 0.

when does that happen?  let's zero out the denominator to check,

x - 2 = 0, thus x = 2

so, if "x" ever becomes 2, the fractions goes kaput.

so, that domain is all real values except that one, that makes the fraction undefined.
6 0
4 years ago
4) Number of rational numbers between – 3 and – 4 are _______ *
wlad13 [49]

Answer:

uncountable

Step-by-step explanation:

----------------uncountable-----------------

8 0
3 years ago
Which expression correctly represents “ten times the difference of a number and six”?
Snowcat [4.5K]

Answer:

I believe that the answer is A

3 0
3 years ago
Read 2 more answers
On a number line, a number, b, is located the same distance from 0 as another number, a, but in the opposite direction. The numb
Komok [63]

Option 1. The equation that represents the direct variation that exists between a and b is b = –a.

<h3>How to find the direct variation here</h3>

The direct variation is of this form y ∝ x

This gives the equation

y = kx

The constant is k

From our question we have

b ∝ a so b = ka

b = 2\frac{3}{4}

a = -2\frac{3}{4}

Remember that b = ka

2\frac{3}{4}  = -2\frac{3}{4} k

This would cancel out to

1 = -k

or k = -1

Hence the equation would be written as

b = -a

Read more on direct variation here:

brainly.com/question/9983911

#SPJ1

3 0
2 years ago
Find the Taylor series for f(x)=sin(x) centered at c=π/2.sin(x)=∑ n=0 [infinity]On what interval is the expansion valid? Give yo
agasfer [191]

Answer:

sinx =1-\frac{1}{2} (x-\frac{\pi}{2} )^2+\frac{1}{24} (x-\frac{\pi}{2} )^4-\frac{1}{720} (x-\frac{\pi}{2} )^6+\frac{1}{40320} (x-\frac{\pi}{2} )^8+...

Step-by-step explanation:

given that f(x) = sin x

we have to find the Taylor series for that

f(x) = sin x   : f( = 1\\f'(x) = cos x :(f'\frac{\pi}{2})=0\\f"(x) = -sinx :f" (\frac{\pi}{2}) =-1\\f^4 (x) = -cosx : f^4 (\frac{\pi}{2}) =0

and so on.

i.e. 2nd, 4th, 6th terms would be 0

and also 1st, 5th, 9th terms would be positive for f value and 3rd, 9th,... would be negative

Using the above we can write Taylor series as

f(x) = f(a)+\frac{f'(a)}{1!} (x-\frac{\pi}{2}) +...+f^n(a) /n! (x- \frac{\pi}{2})^n+...

sinx =1-\frac{1}{2} (x-\frac{\pi}{2} )^2+\frac{1}{24} (x-\frac{\pi}{2} )^4-\frac{1}{720} (x-\frac{\pi}{2} )^6+\frac{1}{40320} (x-\frac{\pi}{2} )^8+...

This is valid for all real values of x.

x ∈(-\infty, infty)

3 0
3 years ago
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