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3 years ago

1. What number, when substituted for x in the following problem, will make the

Mathematics

2 answers:

Sindrei [870]3 years ago

5 0

Answer: 3.33333333333

Step-by-step explanation: 108-88=20

6x=20

6x/6

20/6

x=3.33333333333

Jobisdone [24]3 years ago

3 0

Answer:

x = 3.33333

Step-by-step explanation:

6x + 88 = 108

6x = 108 - 88

6x = 20

x = 20/6

x = 3.33333

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I need help with this question! Will mark brainliest :)

Dvinal [7]

Answer:

y=3x-2

Step-by-step explanation:

you find the 3 (the slope) by seeing how much the y value grows

the -2 is the value of the y-intercept (more clear on a graph) but it's when the point is (0,__) so you subtract 3 from 1.

3 0

3 years ago

Read 2 more answers

Find $\int \dfrac{x}{\sqrt{1-x^4}}$ Please, help

ki77a [65]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2867785

_______________

Evaluate the indefinite integral:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-x^4}}\,dx}\\\\\\ \mathsf{=\displaystyle\int\! \frac{1}{2}\cdot 2\cdot \frac{1}{\sqrt{1-(x^2)^2}}\,dx}\\\\\\ \mathsf{=\displaystyle \frac{1}{2}\int\! \frac{1}{\sqrt{1-(x^2)^2}}\cdot 2x\,dx\qquad\quad(i)}$

Make a trigonometric substitution:

$\begin{array}{lcl}
\mathsf{x^2=sin\,t}&\quad\Rightarrow\quad&\mathsf{2x\,dx=cos\,t\,dt}\\\\
&&\mathsf{t=arcsin(x^2)\,,\qquad 0\ \textless \ x\ \textless \ \frac{\pi}{2}}\end{array}$

so the integral (i) becomes

$\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{1-sin^2\,t}}\cdot cos\,t\,dt\qquad\quad (but~1-sin^2\,t=cos^2\,t)}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{cos^2\,t}}\cdot cos\,t\,dt}$

$\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{cos\,t}\cdot cos\,t\,dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\f dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\,t+C}$

Now, substitute back for t = arcsin(x²), and you finally get the result:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=\frac{1}{2}\,arcsin(x^2)+C}$ ✔

________

You could also make

x² = cos t

and you would get this expression for the integral:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=-\,\frac{1}{2}\,arccos(x^2)+C_2}$ ✔

which is fine, because those two functions have the same derivative, as the difference between them is a constant:

$\mathsf{\dfrac{1}{2}\,arcsin(x^2)-\left(-\dfrac{1}{2}\,arccos(x^2)\right)}\\\\\\
=\mathsf{\dfrac{1}{2}\,arcsin(x^2)+\dfrac{1}{2}\,arccos(x^2)}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \left[\,arcsin(x^2)+arccos(x^2)\right]}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \dfrac{\pi}{2}}$

$\mathsf{=\dfrac{\pi}{4}}$ ✔

and that constant does not interfer in the differentiation process, because the derivative of a constant is zero.

I hope this helps. =)

6 0

3 years ago

A rectangle has sides measuring (2x + 7) units and (5x + 9) units. Part A: What is the expression that represents the area of th

andrey2020 [161]

Answer:

Part A:

(2x+7)(5x+9)

=(2x+7)(5x+9)

=(2x)(5x)+(2x)(9)+(7)(5x)+(7)(9)

=10x2+18x+35x+63

=10x2+53x+63

The formula for determining the area of a rectangle is given as

Area = length × width

Given that the length and width are (2x + 6) units and (5x + 3) units, the expression for the area is

(2x + 6)(5x + 3) = 10x² + 6x + 30x + 18

Area = 10x² + 36x + 18

The degree is 2 because the highest power of the terms is 2. It is classified as a trinomial because it has 3 terms.

C) it is closed under multiplication. the exponents in the polynomials are whole numbers(2 and 1). The whole numbers are closed under addition, which means that the new exponents formed are also whole numbers. The exponents were whole numbers before multiplication and doesn't change after multiplication.

Step-by-step explanation:

3 0

3 years ago

One sandwich is shared equally among eight people. How much will each person get

givi [52]

Answer:

1/8 of a sandwich

3 0

3 years ago

Read 2 more answers

Help me pleasee<br> Questions r in picture

jok3333 [9.3K]

<h2>Answer:</h2>

$\large \bf\implies\frac{12}{61}$

<h2>Step-by-step explanation:</h2>

<h2>Given :</h2>

$\tt \frac{8}{101} , \frac{9}{91} , \frac{10}{81} , \frac{11}{71} ...$

nth term

<h2>Solution :</h2>

We have to add 1 in numerator and -10 in denominator because

$\tt \frac{8}{101} , \frac{9}{91} , \frac{10}{81} , \frac{11}{71} ...[Given]$

$\frac{8 \: + \: 1}{101 \: - \: 10} = \frac{9}{91} \\\\ \frac{9 + 1}{91 - 10} = \frac{10}{81} \\ \\ \frac{10 + 1}{81 - 10} = \frac{11}{71} \\ \\ \frac{11 + 1}{71 - 10} = \frac{12}{61} ...$

The difference is 1 in numerator so we add 1 and the difference is -10 in denominator so we subtract -10.

4 0

2 years ago