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nekit [7.7K]
3 years ago
14

What is the solution to the equation + 5= 8?

Mathematics
2 answers:
evablogger [386]3 years ago
8 0

Answer: 3 + 5 = 8

Step-by-step explanation: if you subtract 5 from 8 (8-5=3) you get 3 so that is your answer

egoroff_w [7]3 years ago
7 0

Answer:

x=13x=13

Step-by-step explanation:

x−5=8

1

 

Add 55 to both sides.

x=8+5x=8+5

2

 

Simplify  8+58+5  to  1313.

x=13x=13

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In cylindrical coordinates, the volume is given with

\displaystyle\int_{\theta=-\pi/2}^{\theta=\pi/2}\int_{r=0}^{r=9}\int_{z=0}^{z=r\cos\theta}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta
=\displaystyle\int_{\theta=-\pi/2}^{\theta=\pi/2}\int_{r=0}^{r=9}r^2\cos\theta\,\mathrm dr\,\mathrm d\theta
=\displaystyle\left(\int_{\theta=-\pi/2}^{\theta=\pi/2}\cos\theta\,\mathrm d\theta\right)\left(\int_{r=0}^{r=9}r^2\,\mathrm dr\right)
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3 years ago
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Solve the equation and find the value of the variable
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The sum of the angles should be 180, so x+ (x+10) + (x+5) = 180. This simplifies to 3x + 15 = 180, so 3x = 165, so x=55.
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3 years ago
A perfume company claims that the mean weight of ther new perfume is at least 8.9 fluid oz. Express the null hypothesis and the
zlopas [31]

Answer:

H0: mu = 8.9 fl oz.\\\\

Ha: mu ≠8.9 fluid oz

Step-by-step explanation:

Given that A perfume company claims that the mean weight of ther new perfume is at least 8.9 fluid oz

For testing this claim, in Statistics we perform a certain measures called hypothesis testing.

For this first step is to create null and alternate hypothesis.

Normally null hypothesis would have some statistic = something

Here we want to test the mean weight of perfume

Hence null hypothesis would be

H0: mu = 8.9 fl oz.

Alternate hypothesis would be opposite of this claim

i.e.

Ha: mu ≠8.9 fluid oz

Hence answer is

H0: mu = 8.9 fl oz.\\\\

Ha: mu ≠8.9 fluid oz

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4 times 2.06 is 8.24 :}

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NO LINKS!!! Part 2: Find the Lateral Area, Total Surface Area, and Volume. Round your answer to two decimal places.​
slava [35]

Answer:

<h3><u>Question 7</u></h3>

<u>Lateral Surface Area</u>

The bases of a triangular prism are the triangles.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the triangles (bases).

\implies \sf L.A.=2(10 \times 6)+(3 \times 6)=138\:\:m^2

<u>Total Surface Area</u>

Area of the isosceles triangle:

\implies \sf A=\dfrac{1}{2}\times base \times height=\dfrac{1}{2}\cdot3 \cdot \sqrt{10^2-1.5^2}=\dfrac{3\sqrt{391}}{4}\:m^2

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<u>Volume</u>

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<h3><u>Question 8</u></h3>

<u>Lateral Surface Area</u>

The bases of a hexagonal prism are the pentagons.

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\sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}a^2

where a is the side length.

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\sf \implies Vol.=area\:of\:base \times height=43.011193... \times 6=258.07\:\:cm^3\:(2\:d.p.)

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