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4 years ago

11

A certain television is advertised as a 41-inch TV (the diagonal length). If the width of the TV is 40 inches, how many inches t

all is the TV?

2 answers:

Bumek [7]4 years ago

6 0

Answer:

9 inches tall

Step-by-step explanation:

Use the Pythagorean theorem: you know the hypotenuse (41) and the leg (40)

$a^{2} +40^{2} =41^{2}$

Solve for a:

Simplify exponents

$a^2+1600=1681$

Subtract 1600 from both sides

$a^2+1600-1600=1681-1600\\a^2=81$

Find the square root

$\sqrt{a^2}=\sqrt{81\\} \\a=9$

KengaRu [80]4 years ago

5 0

Answer:

9 inches

Step-by-step explanation:

A^2+b^2=c^2 so

40^2+b^2=41^2

1600+b^2=1681

so b^2=81

so b=9

You might be interested in

Solve using the elimination method<img src="https://tex.z-dn.net/?f=4x%2B5y%3D4%5C%5C8x%2B10y%3D8" id="TexFormula1" title="4x+5y

EastWind [94]

Answer:

y = all real numbers

x = all real numbers

In set notation/interval notation;

y∈(-∞,∞)

x∈(-∞,∞)

Step-by-step explanation:

1. Background information

The process of elimination involves manipulating an equation such that when the equations in the system are added, one variable is eliminated. Then one can solve for the other variable, finally, one substitutes the answer back in to find the eliminated variable. Therefore solving the system of equations.

2. The given information

Given the system;

(1) 4x + 5y = 4

(2) 8x + 10y = 8

3. Solving

Multiply the first equation by (-2)

4x + 5x = 4

*-2 *-2

-8x - 10x = -8

Add the first and second equations.

-8x - 10y = -8

+ ( 8x + 10y = 8 )

_____________

0

This means that y can be any real number, and hence x can be any real number.

4 0

3 years ago

2 significant figures of 124.4562 L

Charra [1.4K]

Answer:

120 liters.

Step-by-step explanation:

124.4562 rounded to the nearest 2 significant figures is 120.

The third significant figure is 4, which is lower than five, so when rounded don't add +1 to the second significant figure.

The answer is 120 liters.

8 0

3 years ago

Determine whether the integral converges.

Kryger [21]

You have one mistake which occurs when you integrate $\dfrac1{1-p^2}$ . The antiderivative of this is not in terms of $\tan^{-1}p$ . Instead, letting $p=\sin r$ (or $\cos r$ , if you want to bother with more signs) gives $\mathrm dp=\cos r\,\mathrm dr$ , making the indefinite integral equality

$\displaystyle-\frac12\int\frac{\mathrm dp}{1-p^2}=-\frac12\int\frac{\cos r}{1-\sin^2r}\,\mathrm dr=-\frac12\int\sec r\,\mathrm dr=\ln|\sec r+\tan r|+C$

and then compute the definite integral from there.

$-\dfrac12\ln|\sec r+\tan r|\stackrel{r=\sin^{-1}p}=-\dfrac12\ln\left|\dfrac{1+p}{\sqrt{1-p^2}}=\ln\left|\sqrt{\dfrac{1+p}{1-p}}\right|$
$\stackrel{p=u/2}=-\dfrac12\ln\left|\sqrt{\dfrac{1+\frac u2}{1-\frac u2}}\right|=-\dfrac12\ln\left|\sqrt{\dfrac{2+u}{2-u}}\right|$
$\stackrel{u=x+1}=-\dfrac12\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|$
$\implies-\dfrac12\displaystyle\lim_{t\to\infty}\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|\bigg|_{x=2}^{x=t}=-\frac12\left(\ln|-1|-\ln\left|\sqrt{\frac5{-1}}\right|\right)=\dfrac{\ln\sqrt5}2=\dfrac{\ln5}4$

Or, starting from the beginning, you could also have found it slightly more convenient to combine the substitutions in one fell swoop by letting $x+1=2\sec y$ . Then $\mathrm dx=2\sec y\tan y\,\mathrm dy$ , and the integral becomes

$\displaystyle\int_2^\infty\frac{\mathrm dx}{(x+1)^2-4}=\int_{\sec^{-1}(3/2)}^{\pi/2}\frac{2\sec y\tan y}{4\sec^2y-4}\,\mathrm dy$
$\displaystyle=\frac12\int_{\sec^{-1}(3/2)}^{\pi/2}\csc y\,\mathrm dy$
$\displaystyle=-\frac12\ln|\csc y+\cot y|\bigg|_{y=\sec^{-1}(3/2}}^{y=\pi/2}$
$\displaystyle=-\frac12\lim_{t\to\pi/2^-}\ln|\csc y+\cot y|\bigg|_{y=\sec^{-1}(3/2)}^{y=t}$
$\displaystyle=-\frac12\left(\lim_{t\to\pi/2^-}\ln|\csc t+\cot t|-\ln\frac5{\sqrt5}\right)$
$=\dfrac{\ln\sqrt5}2-\dfrac{\ln|1|}2$
$=\dfrac{\ln5}4$

Another way to do this is to notice that the integrand's denominator can be factorized.

$x^2+2x-3=(x+3)(x-1)$

So,

$\dfrac1{x^2+2x-3}=\dfrac1{(x+3)(x-1)}=\dfrac14\left(\dfrac1{x-1}-\dfrac1{x+3}\right)$

There are no discontinuities to worry about since you're integrate over $[2,\infty)$ , so you can proceed with integrating straightaway.

$\displaystyle\int_2^\infty\frac{\mathrm dx}{x^2+2x-3}=\frac14\lim_{t\to\infty}\int_2^t\left(\frac1{x-1}-\frac1{x+3}\right)\,\mathrm dx$
$=\displaystyle\frac14\lim_{t\to\infty}(\ln|x-1|-\ln|x+3|)\bigg|_{x=2}^{x=t}$
$=\displaystyle\frac14\lim_{t\to\infty}\ln\left|\frac{x-1}{x+3}\right|\bigg|_{x=2}^{x=t}$
$=\displaystyle\frac14\left(\lim_{t\to\infty}\ln\left|\frac{t-1}{t+3}\right|-\ln\frac15\right)$
$=\displaystyle\frac14\left(\ln1-\ln\frac15\right)$
$=-\dfrac14\ln\dfrac15=\dfrac{\ln5}4$

Just goes to show there's often more than one way to skin a cat...

7 0

3 years ago

Can you help me??please?!?

enyata [817]

The numbers inside the circles are the number for each one. Where the circles overlap means it is in both.

See attached picture:

8 0

3 years ago

-4(d+5)-3d>8 please help i dont need you to explain it only answer please and thank you

Gala2k [10]

Show your work kidddd

8 0

3 years ago