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2 years ago

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Water boils at 100 \degree \text{C}100°C100, degree, C. This is 400\%400%400, percent more than my room's temperature. What is m

y room's temperature?

2 answers:

Schach [20]2 years ago

7 0

Answer: His room's temperature is 20°C.

Step-by-step explanation:

Since we have given that

Temperature at which water boils = 100°C

According to question, it is 400% more than his room's temperature.

Let his room temperature would be 'x'.

So, our equation becomes,

$\dfrac{100+400}{100}\times x=100\\\\\dfrac{500}{100}\times x=100\\\\5x=100\\\\x=\dfrac{100}{5}\\\\x=20^\circ C$

Hence, his room's temperature is 20°C.

Sophie [7]2 years ago

5 0

Answer:

$20\°C$

Step-by-step explanation:

Let

x------> my room's temperature

we know that

$400\%=\frac{400}{100}=4$

so

The equation that represent this situation is

$4x+x=100$

Solve for x

$5x=100$

$x=20\°C$

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SOLUTION We observe that f '(x) = -1 / (1 + x2) and find the required series by integrating the power series for -1 / (1 + x2).

Ann [662]

Answer:

Required series is:

$\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....$

Step-by-step explanation:

Given that

$f'(x) = -\frac{1}{1 + x^{2}}$ ---(1)

We know that:

$\frac{d}{dx}(tan^{-1}x)=\frac{1}{1+x^{2}}$ ---(2)

Comparing (1) and (2)

$f'(x)=-(tan^{-1}x)$ ---- (3)

Using power series expansion for $tan^{-1}x$

$f'(x)=-tan^{-1}x=-\int {\frac{1}{1+x^{2}} \, dx$

$= -\int{ \sum\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx$

$= -\sum{ \int\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx$

$=-[c+\sum\limits^{ \infty}_{n=0} (-1)^{n}\frac{x^{2n+1}}{2n+1}]$

$=C+\sum\limits^{ \infty}_{n=0} (-1)^{n+1}\frac{x^{2n+1}}{2n+1}$

$=C-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....$

as

$tan^{-1}(0)=0 \implies C=0$

Hence,

$\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....$

7 0

3 years ago

Please please answer this correctly

schepotkina [342]

Answer:

11 m by 18 m

Step-by-step explanation:

The area is the product of two adjacent sides of a rectangle. The perimeter is twice the sum of two adjacent sides, so that sum is (58 m)/2 = 29 m.

We want to find two factors of 198 that sum to 29.

198 = 1·198 = 2·99 = 3·66 = 6·33 = 9·22 = 11·18

Of these factor pairs, only the last one has a sum of 29.

The dimensions of the pool are 11 meter by 18 meters.

6 0

2 years ago

At what point will 3y=4x-6 and 8x-6y=-30 intercept

tester [92]

Answer:

The lines don't intercept

Step-by-step explanation:

we have

$3y=4x-6$

isolate the variable y

Divide by 3 both sides

$y=\frac{4}{3}x-\frac{6}{3}$

simplify

$y=\frac{4}{3}x-2$ -----> equation A

$8x-6y=-30$

Isolate the variable y

$6y=8x+30$

Divide by 6 both sides

$y=\frac{8}{6}x-\frac{30}{6}$

simplify

$y=\frac{4}{3}x-5$ -----> equation B

Compare equation A and equation B

The slopes are the same and the y-intercepts are different

Remember that

If two lines has the same slope, then the lines are parallel

therefore

In this problem line A and line B are parallel lines

The system of equations has no solution, because the lines don't intercept

3 0

3 years ago

(x^3 + 5)(2x – 1)<br>(x^2 – 4)(x^2 + 4)

earnstyle [38]

Answer:

$x^{3} +x^{2} +1$

Step-by-step explanation:

5 0

2 years ago

Solve for x.<br> Please helppp

tankabanditka [31]

Answer:

x = 23

Step-by-step explanation:

2x + 19 + 5x = 180

I say this because it is an angle that is equivalent to the one adjacent to 2x+19, which is on a straight line. That means it is equal to 180.

7x + 19 = 180

180 - 19 = 7x

161/7 = x

x = 23

4 0

2 years ago

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