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3 years ago

Which one is it ? (You cannot click it ) (22 points)!

Mathematics

1 answer:

Sergeu [11.5K]3 years ago

3 0

Answer:

the answer is 0.01 your welcome

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***WILL MAKE BRAINLIEST***

charle [14.2K]

The value of x is 3.

Solution:

Length of the top rectangle = x ft

Width of the top rectangle = x ft

Height of the top rectangle = 3x ft

Volume of the top rectangle $=x \times x\times 3x$

$=3x^{1+1+1}$

Volume of the top rectangle $=3x^3$ cubic feet

Length of the bottom rectangle = $x\sqrt{2}$ ft

Width of the bottom rectangle = $x\sqrt{2}$ ft

Height of the bottom rectangle = 0.5 ft

Volume of the bottom rectangle $=x\sqrt{2} \times x\sqrt{2} \times 0.5$

$=2x^{1+1}\times 0.5$

Volume of the bottom rectangle $=x^2$ cubic feet

Total concrete for both pieces = 90 cubic feet.

Volume of top + volume of bottom = 90 cubic feet

$3 x^{3}+x^{2}=90$

$3 x^{3}+x^{2}-90=0$

Solve by factoring.

$(x-3)\left(3 x^{2}+10 x+30\right)=0$

x = 3, $\left(3 x^{2}+10 x+30\right)=0$

Solve using quadratic formula,

$x=3, x=-\frac{5}{3}+i \frac{\sqrt{65}}{3}, x=-\frac{5}{3}-i \frac{\sqrt{65}}{3}$

Ignore the complex roots.

The value of x is 3.

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4 years ago

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Mice21 [21]

The answer is A hope this helps

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3 years ago

A movie distributor found that the relationship between the amount it spends on advertisements for a particular movie and the am

icang [17]

Answer:

i believe the answer is B, im not 100% though

Step-by-step explanation:

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3 years ago

Is 5.333333 rational or irrational

tekilochka [14]

Step-by-step explanation:

irrational ok????>>>>

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3 years ago

Read 2 more answers

Please help math questions put number with answers please

natita [175]

Answer:

See below for answers

Step-by-step explanation:

Problem 1

$\frac{1}{sin\theta}=2cos\theta\: ; \: 0\leq\theta < 2\pi\\\\1=2sin\theta cos\theta\\\\1=sin2\theta\\\\\frac{\pi}{2}+2\pi n=2\theta\\ \\\frac{\pi}{4}+\pi n=\theta\\ \\\theta=\bigr\{\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\bigr\}$

Therefore, the second option is correct.

Problem 2

$cos(2x)+1=sin(x)+2\:;\: 0\leq x < 2\pi\\\\1-2sin^2x+1=sin(x)+2\\\\-2sin^2x=sin(x)\\\\0=2sin^2x+sinx\\\\0=sinx(2sinx+1)$

$0=sinx\\\\x=\pi n\\\\x=0,\pi$

$0=2sinx+1\\\\-1=2sinx\\\\-\frac{1}{2}=sinx\\ \\x=\frac{7\pi}{6}+2\pi n,\frac{11\pi}{6}+2\pi n\\\\x=\frac{7\pi}{6},\frac{11\pi}{6}$

Therefore, the solution set is $x=\bigr\{0,\pi,\frac{7\pi}{6},\frac{11\pi}{6}\bigr\}$ , making the second option correct.

Problem 3

$2cos^2x=-cosx\: ;\: 0^\circ\leq x < 360^\circ\\\\2cos^2x+cosx=0\\\\cosx(2cosx+1)=0$

$cosx=0\\\\x=\frac{\pi}{2}+\pi n\\ \\x=90^\circ+180n^\circ\\\\x=90^\circ,270^\circ$

$2cosx+1=0\\\\2cosx=-1\\\\cosx=-\frac{1}{2}\\\\x=\frac{2\pi}{3}+2\pi n,\frac{4\pi}{3}+2\pi n\\ \\x=120^\circ+360n^\circ,240^\circ+360n^\circ\\\\x=120^\circ,240^\circ$

Therefore, the solution set is $x=\bigr\{90^\circ,120^\circ,240^\circ,270^\circ\bigr\}$ , making the fourth option correct

7 0

2 years ago