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3 years ago

Work out the surface area of the solid cuboid L=5cm W=2cm H=5.3cm

Mathematics

1 answer:

Ludmilka [50]3 years ago

3 0

Answer:

A = 94.2 cm²

Step-by-step explanation:

Given that,

Length, l = 5 cm

Width, b = 2 cm

Height, h = 5.3 cm

We need to find the surface area of the solid cuboid.

The formula for the surface area of the sold cubiod is given by :

A = 2(lb+bh+hl)

Substitute the values in the above formu;a

A = 2(5(2)+2(5.3)+(5.3)(5))

= 2(10+10.6+26.5)

= 2(47.1)

= 94.2 cm²

Hence, the required surface area is 94.2 cm².

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This organelle is responsible for controlling the cell's activities. 1.Mitochondria 2.Ribosome 3.Nucleus 4.Endoplasmic Reticulum

Inga [223]

Hi there,

The Mitochondria is the power house of the cell. It provides energy for the cell therefore that is out.

The Ribosome breaks down and produces protein for the cell therefore that is out.

The Endoplasmic Reticulum allows protein to pass through in vacuoles therefore that's out as well.

The Nucleus acts as a brain controlling other organelles therefore it's the Nucleus.

So the answer is 3.

Hope this helps!

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3 years ago

1. Sand falling from a chute forms a conical pile whose altitude is always equal to 4/3 the radius of the base. (a) How fast is

solong [7]

Answer:

The volume is increasing at a rate of 16286 in³/min.

Step-by-step explanation:

a) The volume of a cone is given by:

$V = \frac{1}{3}h\pi r^{2}$

Where:

r: is the radius

h: is the height

The rate of change of the volume can be calculated by using the chain rule:

$\frac{dV}{dt} = \frac{\pi}{3}[\frac{dh}{dt}r^{2} + h\frac{d(r^{2})}{dt}]$

Since h = 4/3 r we have:

$\frac{dV}{dt} = \frac{\pi}{3}[\frac{d(\frac{4r}{3})}{dt}r^{2} + \frac{4r}{3}\frac{d(r^{2})}{dt}]$

$\frac{dV}{dt} = \frac{4\pi}{9}[\frac{dr}{dt}r^{2} + r\frac{d(r^{2})}{dt}]$

$\frac{dV}{dt} = \frac{4\pi}{9}[\frac{dr}{dt}r^{2} + 2r^{2}\frac{dr}{dt}]$ (1)

With:

$\frac{dr}{dt}$ = 3 in/min

$r = 3 ft*\frac{12 in}{1 ft} = 36 in$

And by entering the above values into equation (1) we have:

$\frac{dV}{dt} = \frac{4\pi}{9}[(3 in/min)*(36 in)^{2} + 2*(36 in)^{2}*3 in/min]$

$\frac{dV}{dt} = 16286 in^{3}/min$

Therefore, the volume is increasing at a rate of 16286 in³/min.

I hope it helps you!

3 0

3 years ago

36. Which side of the triangle in the diagram is the hypotenuse? A X B. Y C. Z

Nikolay [14]

Answer:

C) Z

Step-by-step explanation:

In the right triangle, X and Y are side lengths, while Z is the hypotenuse, so C is the correct answer.

6 0

3 years ago

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-1 _ * 8 6 not -1/6 it is negative 1 over 6

RSB [31]

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3 years ago

It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per nig

Kitty [74]

Answer: No , at 0.05 level of significance , we have sufficient evidence to reject the claim that LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average.

Step-by-step explanation:

Let $\mu$ denotes the average hours of sleep per night.

As per given , we have

$H_0:\mu=7\\H_a:\mu$

, since $H_a$ is left-tailed and population standard deviation is unknown, so the test is a left-tailed t -test.

Also , it is given that ,

Sample size : n= 22

Sample mean : $\overline{x}=7.24$

Sample standard deviation : s= 1.93

Test statistic : $t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}$

i.e. $t=\dfrac{7.24-7}{\dfrac{1.93}{\sqrt{22}}}\approx0.58$

For significance level $\alpha=0.05$ and degree of freedom 21 (df=n-1),

Critical t-value for left-tailed test= $t_{\alpha, df}=t_{0.05,21}=- 1.7207$

Decision : Since the test statistic value (0.58) > critical value 1.7207, it means we are failed to reject the null hypothesis .

[Note : When $|t_{cal}|>|t_{cri}|$ , then we accept the null hypothesis.]

Conclusion: We have sufficient evidence to reject the claim that LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average.

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3 years ago