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3 years ago

Solve the equation. You will not get credit if you do not show your work or 4 points

Mathematics

1 answer:

Feliz [49]3 years ago

5 0

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CAN SOMEONE PLEASE HELP ME WITH MY HOMEWORK ILL DO ANYTHING FOR HELP JUST PLEASE HELP ME WITH MY HOMEWORK AND PLEASE EXPLAIN HOW

icang [17]

Answer:

the simplest way to finish your hw is to ask the Q of your hw here instead of asking us to help with your hw

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3 years ago

Read 2 more answers

Mrs. Byrne's class went raspberry picking. The data show the weights of the cartoons of raspberries the students picked. Make a

Licemer1 [7]

Here is what you tally chart/table would look like:

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3 years ago

Which of the following expressions is equivalent to 32 ÷ 4• 4 ÷ 2 • 2?

mel-nik [20]

Answer:

128

Step-by-step explanation:

32 ÷ 4• 4 ÷ 2 • 2

8 .4.2.2

128

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3 years ago

1. -88 = -3(4m + 5) - (1 - 3m)

Vadim26 [7]

Answer:

m=8

Step-by-step explanation:

-88=-3(4m+5)-(1-3m)

-88=-12m-15-(1-3m) <- Distributive Property

-88=-12m-15-1+3m <- Open () if there is a negative negative the symbol equals positive

-88=-9m-16 <- Simplify

0=-9m+72 <- Add 88 to both sides

9m = 72 <- Add 9m to both sides

9m = 72

/9 /9

m=8

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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