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2 years ago

Solve the following equation for x, in terms of a and b. ax = 15 + bx

Mathematics

1 answer:

charle [14.2K]2 years ago

8 0

Value of x will be $x=\frac{15}{a-b}$ .

Equation given in the question → $ax=15+bx$

To find the value of 'x' in terms of a and b, we will simplify the given equation,

$ax=15+bx$

$ax-bx=(15+bx)-bx$

$x(a-b)=15$

$x=\frac{15}{a-b}$

Therefore, Values to filled → Blank 1 = 15

Blank 2 = (a - b)

Learn more,

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The ratio is 8:9 is the ratio between 0.45 - 0.60?

bagirrra123 [75]

Divide 8 by 9

8 / 9 = 0.888

0.88 is greater than 0.60 so it is not between 0.45 - 0.60

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3 years ago

Square root of -1? PLEASE HELP

kolezko [41]

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3 years ago

12.20h - 7.15 = 90.45

zhannawk [14.2K]

Answer:

H=8

Step-by-step explanation:

12.2H-7.15=90.45

12.2H=90.45+7.15

12.2H=97.6

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6 0

2 years ago

Read 2 more answers

Pls help I don't understand:

sineoko [7]

Answer:

Part a) $T(d)=2d+30$

Part b) $T(6)=42\ minutes$

Step-by-step explanation:

Part a) Write an equation for T (d)

Let

d ----> the number of days

T ---> the time in minutes of the treadmill

we know that

The linear equation in slope intercept form is equal to

$T=md+b$

where

m is the slope or unit rate

b is the y-intercept or initial value

In this problem we have

The slope or unit rate is

$m=2\ \frac{minutes}{day}$

The y-intercept or initial value is

$b=30\ minutes$

substitute

$T(d)=2d+30$

Part b) Find T (6), the minutes he will spend on the treadmill on day 6

For d=6

substitute in the equation and solve for T

$T(6)=2(6)+30$

$T(6)=42\ minutes$

8 0

3 years ago

Solve only if you know the solution and show work.

SashulF [63]

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=\int\mathrm dx+2\int\frac{\sin x+3}{\cos x+\sin x+1}\,\mathrm dx$

For the remaining integral, let $t=\tan\dfrac x2$ . Then

$\sin x=\sin\left(2\times\dfrac x2\right)=2\sin\dfrac x2\cos\dfrac x2=\dfrac{2t}{1+t^2}$
$\cos x=\cos\left(2\times\dfrac x2\right)=\cos^2\dfrac x2-\sin^2\dfrac x2=\dfrac{1-t^2}{1+t^2}$

and

$\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx\implies \mathrm dx=2\cos^2\dfrac x2\,\mathrm dt=\dfrac2{1+t^2}\,\mathrm dt$

Now the integral is

$\displaystyle\int\mathrm dx+2\int\frac{\dfrac{2t}{1+t^2}+3}{\dfrac{1-t^2}{1+t^2}+\dfrac{2t}{1+t^2}+1}\times\frac2{1+t^2}\,\mathrm dt$

The first integral is trivial, so we'll focus on the latter one. You have

$\displaystyle2\int\frac{2t+3(1+t^2)}{(1-t^2+2t+1+t^2)(1+t^2)}\,\mathrm dt=2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt$

Decompose the integrand into partial fractions:

$\dfrac{3t^2+2t+3}{(1+t)(1+t^2)}=\dfrac2{1+t}+\dfrac{1+t}{1+t^2}$

so you have

$\displaystyle2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt=4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$

which are all standard integrals. You end up with

$\displaystyle\int\mathrm dx+4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$
$=x+4\ln|1+t|+2\arctan t+\ln(1+t^2)+C$
$=x+4\ln\left|1+\tan\dfrac x2\right|+2\arctan\left(\arctan\dfrac x2\right)+\ln\left(1+\tan^2\dfrac x2\right)+C$
$=2x+4\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)+C$

To try to get the terms to match up with the available answers, let's add and subtract $\ln\left|1+\tan\dfrac x2\right|$ to get

$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)-\ln\left|1+\tan\dfrac x2\right|+C$
$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|+C$

which suggests A may be the answer. To make sure this is the case, show that

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\sin x+\cos x+1$

You have

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\cos^2\dfrac x2+\sin\dfrac x2\cos\dfrac x2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\dfrac{1+\cos x}2+\dfrac{\sin x}2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac2{\cos x+\sin x+1}$

So in the corresponding term of the antiderivative, you get

$\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|=\ln\left|\dfrac2{\cos x+\sin x+1}\right|$
$=\ln2-\ln|\cos x+\sin x+1|$

The $\ln2$ term gets absorbed into the general constant, and so the antiderivative is indeed given by A,

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=2x+5\ln\left|1+\tan\dfrac x2\right|-\ln|\cos x+\sin x+1|+C$

5 0

2 years ago