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4 years ago

11

Find the exact length of the third side

1 answer:

Paha777 [63]4 years ago

7 0

Answer:

a = √58

Step-by-step explanation:

a² + b² = c²

Fill in the values we know:

a² + 5² = √83²

a² + 25 = 83

Subtract 25 from both sides:

a² + 25 - 25 = 83 - 25

a² = 58

Square root both sides to get 'a' by itself:

√a² = √58

a = √58

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3x+15/ 6-x for what value of x is the rational expression below to equal zero

Alex73 [517]

X = -5.

You need the numerator to equal 0 in order for the expression to equal 0. So set the top equal to 0 and solve.

5 0

3 years ago

Read 2 more answers

What is the equation of the line that passes through the point (6,-6) and is parallel to the x axis

Rzqust [24]

Answer:

$y=-6$

Step-by-step explanation:

we know that

A line parallel to the x-axis is a horizontal line

The equation of a horizontal line is equal to the y-coordinate of the point that passes through it

In this problem the line passes through the point (6,-6)

so

The y-coordinate is -6

therefore

The equation of the line is

$y=-6$

4 0

4 years ago

2 points) Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One

Stells [14]

$y'=(t+y)^2-1$

Substitute $u=t+y$ , so that $u'=y'$ , and

$u'=u^2-1$

which is separable as

$\dfrac{u'}{u^2-1}=1$

Integrate both sides with respect to $t$ . For the integral on the left, first split into partial fractions:

$\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1$

$\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt$

$\dfrac12(\ln|u-1|-\ln|u+1|)=t+C$

Solve for $u$ :

$\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C$

$\ln\left|1-\dfrac2{u+1}\right|=2t+C$

$1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}$

$\dfrac2{u+1}=1-Ce^{2t}$

$\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}$

$u=\dfrac2{1-Ce^{2t}}-1$

Replace $u$ and solve for $y$ :

$t+y=\dfrac2{1-Ce^{2t}}-1$

$y=\dfrac2{1-Ce^{2t}}-1-t$

Now use the given initial condition to solve for $C$ :

$y(3)=4\implies4=\dfrac2{1-Ce^6}-1-3\implies C=\dfrac3{4e^6}$

so that the particular solution is

$y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}$

3 0

3 years ago

If jackeline and Raychel decide to watch only the two longest movies and take a 30 minute break in between how much of their fiv

Allushta [10]

Why are their names spelt like that? It’s jaclyn & Rachel
Not JAISBSKSJ
AND RAISBSMAJSBS

7 0

3 years ago

Thanks for the help!

NikAS [45]

Answer:

A

Step-by-step explanation:

So we have the piecewise function:

$f(x)= \begin{cases}\sqrt{2x} & \text{if }x$

And we want to find f(8).

Since our input value is 8, choose the equation that fits our input.

The first equation demand x to be less than 3. 8 is not less than 3, so we won't use that.

The second equation demand x to e greater than or equal to 3 and less than 8. 8 is <em>not</em> less than 8. So, we won't use that.

The third equation demands x to be greater than or equal to 8. 8 is greater than or equal to 8, so we'll use the third equation.

So:

$f(x)=42\text{ if }x\geq 8$

$f(8)=42$

There're nothing more to do, we're done :)

The answer is A.

Edit: Improved Format

3 0

4 years ago