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4 years ago

Neeeeeed the help please

Mathematics

1 answer:

lions [1.4K]4 years ago

8 0

The correct answer is -5/7. Hope this helps.

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There is 1 inch of snow on the ground when it begins to snow 0.75 inch per hour. Which linear equation represents the total dept

aleksandrvk [35]

Y=0.75x+1 is the answer

3 0

4 years ago

I need to know what x equals. HELP!!!!

SSSSS [86.1K]

Answer:

X=4

Step-by-step explanation:

4*4=16+7=23

8 0

3 years ago

Read 2 more answers

Place 3 numbers between 12 and 60 to make a sequence of 5 numbers with a common difference

Vilka [71]

The 3 numbers which should be placed between 12 and 60 are; 24, 36 and 48.

<h3>What three numbers should be placed between 12 and 60?</h3>

It follows from the task content that the first and last numbers are; 12 and 60 respectively.

Hence, since the difference between 12 and 60 is 48 and there are 4 transitions between the two numbers to have 5 total numbers,

It follows that the common difference is; 48/4 = 12 and the three numbers required are; 24,36 and 48.

Read more on common difference;

brainly.com/question/19601308

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5 0

2 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

Solve for x D) 10y=130

mixer [17]

X equals 13 and that is the answer

5 0

3 years ago