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3 years ago

Please help...no links please!

Mathematics

1 answer:

Illusion [34]3 years ago

6 0

Answer:

Step-by-step explanation:

Sides AB and sides EF have no connection to each other, therefore, they would not contribute to the statement that both triangles are the same.

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Sheila purchased ski equipment for $513 using a six-month deferred payment plan. The interest rate after the introductory period

klemol [59]

$65 * 6 = $390 total monthly payments made

The downpayment made is $125, so the total balance at the end is:

balance = $513 – ($125 + $390)

balance = - $2

There is an excess of 2 dollars at the end.

4 0

3 years ago

the sum of two numbers is 57. if the larger of the two numbers is 9 less than twice the smaller, find the two numbers

svp [43]

Answer:

The answer is 35 and 22

Because:

U let the numbers(#) be x and y (x will be the larger number and y the smaller number

So:

x + y =57

x = 2*y - 9

Therefore,

{2*y - 9} + y = 57

Leaving the two numbers 35 and 22

I hope I could help

7 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

Write the equation of each line using the given information.

Leni [432]

a) x – 2y + 6 = 0 b) x + y = 1 c) y = 1 d) y = -3x + 8

<h3>Solution:</h3>

a. The points (-4,1) and (2,4) both lie on the line

The general line equation on which (a, b) and (c, d) lies is:

$y-\mathrm{b}=\frac{d-b}{c-a}(x - a)$

Here the given points are (a, b) = (-4, 1) and (c, d) = (2, 4)

Thus the required equation is:

$y-1=\frac{4-1}{2-(-4)}(x-(-4))$

On solving we get,

$\begin{array}{l}{\rightarrow y-1=\frac{3}{2+4}(x+4)} \\\\ {\rightarrow y-1=\frac{3}{6}(x+4)} \\\\ {\rightarrow 2(y-1)=1(x+4)} \\\\ {\rightarrow 2 y-2=x+4} \\\\ {\rightarrow x-2 y+6=0}\end{array}$

b.) m= -1 and the point (2, -1) lies on the line

The equation of line in point slope form is y – b = m(x – a)

where m is slope and (a, b) is a point on it

Here m = -1 and (a, b) = (2, -1)

Thus the required equation is:

y – (-1) = -1(x - 2)

y + 1 = -x + 2

y = -x + 2 -1

y = -x + 1

c. )It has the same slope as y = 5 and passes through (1, 1)

our line has same slope with y = 5, then our equation would be y = k and it passes through (x, y) = (1, 1) so, then by substitution

1 = k

k =1

Then our equation will be y = k

y = 1

d. ) m= -3 and it has a y-intercept of (0, 8)

line equation in slope intercept form is y = mx + b where m is slope and b is y – intercept.

Then, our equation will be y = -3x + 8

We took y- intercept = 8 as it is the value of y when x = 0

6 0

3 years ago

Simplify 2.5 (16/2-2.3) - 2

dsp73

Answer:

it's just ( -3.3 )but I need to put 20 + words and stuff so read this if you would like

7 0

2 years ago