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3 years ago

Why is 1 + (−5) equal to −4?

Mathematics

2 answers:

ruslelena [56]3 years ago

8 0

Answer:

because its 5 units to the left on a horizontal number line

tester [92]3 years ago

8 0

Answer:

The answer to the question is D

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Use trigonometric substitution to evaluate the integral 13 + 12x − x2 dx . First, write the expression under the radical in an a

SCORPION-xisa [38]

I don't see a square root sign anywhere, so I'll assume the integral is

$\displaystyle\int\sqrt{13+12x-x^2}\,\mathrm dx$

First complete the square:

$13+12x-x^2=49-(6-x)^2=7^2-(6-x)^2$

Now in the integral, substitute

$6-x=7\sin t\implies\mathrm dx=-7\cos t\,\mathrm dt$

so that

$t=\sin^{-1}\left(\dfrac{6-x}7\right)$

Under this change of variables, we have

$7^2-(6-x)^2=7^2-7^2\sin^2t=7^2(1-\sin^2t)=7^2\cos^2t$

so that

$\displaystyle\int\sqrt{13+12x-x^2}\,\mathrm dx=-7\int\sqrt{7^2\cos^2t}\,\cos t\,\mathrm dt=-49\int|\cos t|\cos t\,\mathrm dt$

Under the right conditions, namely that cos(<em>t</em>) > 0, we can further reduce the integrand to

$|\cos t|\cos t=\cos^2t=\dfrac{1+\cos(2t)}2$

$\displaystyle-49\int|\cos t|\cos t\,\mathrm dt=-\frac{49}2\int(1+\cos(2t))\,\mathrm dt=-\frac{49}2\left(t+\frac12\sin(2t)\right)+C$

Expand the sine term as

$\dfrac12\sin(2t)}=\sin t\cos t$

Then

$t=\sin^{-1}\left(\dfrac{6-x}7\right)\implies \sin t=\dfrac{6-x}7$

$t=\sin^{-1}\left(\dfrac{6-x}7\right)\implies \cos t=\sqrt{7^2-(6-x)^2}=\sqrt{13+12x-x^2}$

So the integral is

$\displaystyle-\frac{49}2\left(\sin^{-1}\left(\dfrac{6-x}7\right)+\dfrac{6-x}7\sqrt{13+12x-x^2}\right)+C$

3 0

2 years ago

A nearby school for wizardry is larger. They have 14 times as many students.

RoseWind [281]

Answer:

14 times more than the larger school

Step-by-step explanation: This makes no sense, there is no problem to solve, just a statement

7 0

2 years ago

Identify the diameter of the disc

Olegator [25]

radius = (4*10^2 + 24^2)/8*10 =

(400 + 576)/80=

976/80 = 12.2

diameter = 12.2 x 2 = 24.4

4 0

2 years ago

What is 5/6 divided by-2/3

liraira [26]

Answer: -5/4

Step-by-step explanation:

Apply the fraction rule: a/-b = -a/b

-5/6/2/3

-----------------------------------------------------------------------------------------------------------------

Divide fractions: a/b/c/d = a x d/ b x c

-5 x 3 / 6 x 2

-----------------------------------------------------------------------------------------------------------------

Multiply the numbers: 6 x 2 = 12

-5 x 3 / 12

-----------------------------------------------------------------------------------------------------------------

Multiply the numbers: 5 x 3 = 15

-15/12

-----------------------------------------------------------------------------------------------------------------

Cancel the common factor.3

-5/4

-----------------------------------------------------------------------------------------------------------------

Sorry if this is a little confusing but your answer to the question is ( -5/4 )

3 0

3 years ago

Find a function where f(0)=2 and f(1)=2

xenn [34]

Answer:

Do you want to be extremely boring?

Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?

$f(x) = 2$ is a valid solution.

Want something more fun? Why not a parabola? $f(x)= ax^2+bx+c$ .

At this point you have three parameters to play with, and from the fact that $f(0)=2$ we can already fix one of them, in particular $c=2$ . At this point I would recommend picking an easy value for one of the two, let's say $a= 1$ (or even $a=-1$ , it will just flip everything upside down) and find out b accordingly: $f(1)=2 \rightarrow 1^2+b+2=2 \rightarrow b=-1$

Our function becomes

$f(x) = x^2-x+2$

Notice that it works even by switching sign in the first two terms: $f(x) = -x^2+x+2$

Want something even more creative? Try playing with a cosine tweaking it's amplitude and frequency so that it's period goes to 1 and it's amplitude gets to 2: $f(x) = A cos (kx)$

Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need $A= 2$ , and at that point the first condition is guaranteed; using the second to find k we get $2= 2 cos (k1) = cos k = 1 \rightarrow k = 2\pi$

$f(x) = 2cos(2\pi x)$

Or how about a sine wave that oscillates around 2? with a similar reasoning you get

$f(x)= 2+sin(2\pi x)$

Sky is the limit.

8 0

2 years ago