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2 years ago

Solve for r in the scientific formula d = rt. r = _______

Mathematics

2 answers:

BartSMP [9]2 years ago

5 0

D=rt.

Get r by itself. So, divide both sides by 1.

d/t= r

I hope this helps!
~kaikers

marta [7]2 years ago

4 0

R = d/t is the answer. Hope this helped

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What is the oblique asymptote of the function f(x) x^2+x-2/x+1

Mamont248 [21]

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

$\displaystyle\lim_{x\to\infty}\left\{f(x)-x \right\}=\lim_{x\to\infty}\frac{2}{x+1}=\lim_{x\to\infty}\frac{\displaystyle\frac{2}{x}}{1+\displaystyle\frac{1}{x}}
=0$

Consequently, the limit of $f(x)$ as x approaches infinity is $x$ .

In other words, $f(x)$ approaches the line y=x,

so oblique asymptote is y=x.

I'm Japanese, if you find some mistakes in my English, please let me know.

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3 years ago

Write 3.554 as a percent

tatiyna

Step-by-step explanation:

0.03554%.............

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2 years ago

Simplify to an expression of the form (a sin(θ)). 2sin(/8)2cos(/8)

rjkz [21]

Answer:

Your expression equals $2 \sin(\frac{\pi}{4})$ assuming the problem was to simplify $2\sin(\frac{\pi}{8}) 2\cos(\frac{\pi}{8})$ to the form $a\sin(\theta)$ .

Step-by-step explanation:

$2\sin(\frac{\pi}{8}) 2\cos(\frac{\pi}{8})$

$2(2)\sin(\frac{\pi}{8})\cos(\frac{\pi}{8})$

Let's use the following identity: $2\sin(A)\cos(A)=\sin(2A)$

$2\sin(2 \cdot \frac{\pi}{8})$

$2 \sin(\frac{\pi}{4})$ which is comparable to the form $a\sin(\theta)$ .

$a=2$

$\theta=\frac{\pi}{4}$

5 0

2 years ago

19% of 43 estimated answer

Phantasy [73]

Answer:

19% of 43 = 8.17

7 0

3 years ago

Read 2 more answers

Try this hard Math Problem if you dare!!

Dennis_Churaev [7]

Answer:

a. $(x - 3)^2 + 16$

b. $8(x -7)^2$

c. $(a^2 - 1)(7x - 6)$ or $(a+1)(a-1)(7x-6)$

d. $(x^2-4)(x^2+3)$ or $(x-2)(x+2)(x^2+3)$

e. $(a^n+b^n)(a^n-b^n)(a^{2n} +b^{2n})$

Step-by-step explanation:

$a.\ (x + 1)^2 - 8(x - 1) + 16$

Expand

$(x + 1)(x + 1) - 8(x - 1) + 16$

Open brackets

$x^2 + x + x + 1 - 8x + 8 + 16$

$x^2 + 2x + 1 - 8x + 24$

Collect Like Terms

$x^2 + 2x - 8x+ 1 + 24$

$x^2 - 6x+ 25$

Express 25 as 9 + 16

$x^2 - 6x+ 9 + 16$

Factorize:

$x^2 - 3x - 3x + 9 + 16$

$x(x -3)-3(x - 3) + 16$

$(x - 3)(x - 3) + 16$

$(x - 3)^2 + 16$

$b.\ 8(x - 3)^2 - 64(x-3) + 128$

Expand

$8(x - 3)(x - 3) - 64(x-3) + 128$

$8(x^2 - 6x+ 9) - 64(x-3) + 128$

Open Brackets

$8x^2 - 48x+ 72 - 64x+192 + 128$

Collect Like Terms

$8x^2 - 48x - 64x+192 + 128+ 72$

$8x^2 -112x+392$

Factorize

$8(x^2 -14x+49)$

Expand the expression in bracket

$8(x^2 -7x-7x+49)$

Factorize:

$8(x(x -7)-7(x-7))$

$8((x -7)(x-7))$

$8(x -7)^2$

$c.\ 7a^2x - 6a^2 - 7x + 6$

Factorize

$a^2(7x - 6) -1( 7x - 6)$

$(a^2 - 1)(7x - 6)$

The answer can be in this form of further expanded as follows:

$(a^2 - 1^2)(7x - 6)$

Apply difference of two squares

$(a+1)(a-1)(7x-6)$

$d.\ x^4 - x^2 - 12$

Express $x^4$ as $x^2$

$(x^2)^2 - x^2 - 12$

Expand

$(x^2)^2 +3x^2- 4x^2 - 12$

$x^2(x^2+3) -4(x^2+3)$

$(x^2-4)(x^2+3)$

The answer can be in this form of further expanded as follows:

$(x^2-2^2)(x^2+3)$

Apply difference of two squares

$(x-2)(x+2)(x^2+3)$

$e.\ a^{4n} -b^{4n}$

Represent as squares

$(a^{2n})^2 -(b^{2n})^2$

Apply difference of two squares

$(a^{2n} -b^{2n})(a^{2n} +b^{2n})$

Represent as squares

$((a^{n})^2 -(b^{n})^2)(a^{2n} +b^{2n})$

Apply difference of two squares

$(a^n+b^n)(a^n-b^n)(a^{2n} +b^{2n})$

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3 years ago