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3 years ago

Write a situation that can be represented by the number-42

Mathematics

1 answer:

Inessa [10]3 years ago

3 0

Answer:

In a real life situation you can represent -42 by saying your bank account has -42 dollars. Meaning that you owe money to the bank.

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What is the value of x in the equation below? 1+2e^x+1=9

olga2289 [7]

I am sure the correct answer is x=0.38629436…hope this help you

8 0

3 years ago

Read 2 more answers

PLEASE HELP! WILL GIVE 70 POINTS!!

Tresset [83]

Answer:

$A=189\ mm^2$

Step-by-step explanation:

Surface Areas

Is the sum of all the lateral areas of a given solid. We need to compute the total surface area of the given prism. It has 5 sides, two of them are equal (top and bottom areas) and the rest are rectangles.

Computing the top and bottom areas. They form a right triangle whose legs are 4.5 mm and 6 mm. The area of both triangles is

$\displaystyle A_t=2*\frac{b.h}{2}=b.h=(4.5)(6)=27 mm^2$

The front area is a rectangle of dimensions 7.7 mm and 9 mm, thus

$A_f=b.h=(7.5)(9)=67.5 \ mm^2$

The back left area is another rectangle of 4.5 mm by 9 mm

$A_l=b.h=(4.5)(9)=40.5 \ mm^2$

Finally, the back right area is a rectangle of 6 mm by 9 mm

$A_r=b.h=(6)(9)=54 \ mm^2$

Thus, the total surface area of the prism is

$A=A_t+A_f+A_l+A_r=27+67.5+40.5+54=189\ mm^2$

$\boxed{A=189\ mm^2}$

4 0

3 years ago

Pls show full working out

sweet-ann [11.9K]

Answer:

Question 1a. i) $x=1.8m$

Question 1a. ii) $Volume=27.6m^3$

Question 1b) $Volume=65.9m^3$

Explanation:

Question 1 a. i) Find the value of x.

$tan(\theta )=\dfrac{opposite\text{ }leg}{adjacent\text{ }leg}$

For the smalll triangle you can write:

$tan(\theta )=\dfrac{x}{1m}$

For tthe big triangle:

$tan(\theta )=\dfrac{x+2.7m}{2.5m}$

Substitute:

$\dfrac{x}{1m}=\dfrac{x+2.7m}{2.5m}$

Solve for x:

$2.5x=x+2.7m\\\\2.5x-x=2.7m\\\\1.5x=2.7m\\\\x=2.7m/1.5\\\\x=1.8m$

Question 1a ii) Find the volume of the frustrum

Find the volume of a cone with height = 2.7m + 1.8m = 4.5m, and radius = 2.5m

Formula:

$Volume=(1/3)\pi \times radius^2\times height$

Substitute:

$Volume=(1/3)\pi \times (2.5m)^2\times 4.5m=9.375\pi m^3$

Find the volume of a cone with heigth = 1.8m and radius = 1m

$Volume=(1/3)\pi \times (1m)^2\times 1.8m=0.6\pi m^3$

Subtract the volume of the small cone from the volume of the big cone

$Volume\text{ }of\text{ }frustrum=9.375\pi m^3-0.6\pi m^3=8.775\pi m^3\approx 27.6m^3$

Question 1b. Calculate the volume of the bin

i) Upper frustrum

This is the same frustrum from the equation of above, thus ist volume is 27.6m³.

ii) Lower frustrum

$\dfrac{x}{2.0m}=\dfrac{x+2.4m}{2.5m}$

$2.5x=2(x+2.4m)\\\\2.5x=2x+4.8m\\\\0.5x=4.8m\\\\x=9.6m$

$Volume=(1/3)\pi \times (2.5m)^2\times (9.6m+2.4m)-(2.0m)^2\times (9.6m)$

$Volume=38.3m^3$

iii) Add the volume of the two frustrums

$Volume=27.6m^3+38.3m^3=65.9m^3$

6 0

3 years ago

⚠️URGENT PLS HELP LINKS WILL BE REPORTED

Arte-miy333 [17]

Answer:

e and c

Step-by-step explanation:

4 0

3 years ago

Let R be the shaded region in the first quadrant enclosed by the y-axis and the graphs of y=sinx and y=cosx.

julsineya [31]

since sin and cos = each other at pi/4; take your integrals from 0 to pi/4
[S] cos(t) dt - [S] sin(t) dt ;[0,pi/4]

to revolve it around the x axis;
we do a sum of areas [S] 2pi [f(x)]^2 dx

take the cos first and subtract out the sin next; like cutting a hole out of a donuts.
pi [S] cos(x)^2 dx - [S] sin(x)^2 dx ; [0,pi/4]

cos(2t) = 2cos^2 - 1
cos^2 = (1+cos(2t))/2

1/sqrt(2) - (-1/sqrt(2) +1)
1/sqrt(2) + 1/sqrt(2) -1
(2sqrt(2) - sqrt(2))/sqrt(2) = sqrt(2)/sqrt(2) = 1

3 0

3 years ago