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2 years ago

10

Which number line shows 485+408?

1 answer:

arsen [322]2 years ago

5 0

What are the number lines

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Urgent help! This is very important please help me fast! How do you do: Expand & simplify:(x 4)(x 2)

ahrayia [7]

So the question is asking: (x-4)(x-2)
because there is a bracket in between the two expressions it means we have to multiply them together: (x-4) x (x-2)

you can break down the question into smaller parts:
x multiply x
x multiply -2
-4 multiply x
-4 multiply -2

here are the answers:
x^2 (means x squared)
-2x
-4x
8 (because when you multiply two negative numbers it makes a positive)

now you put it into an expression (this is expanding):
x^2 - 2x - 4x + 8

to simplify it you collect like terms:
x^2 - 6x + 8

The above is the answer :)

4 0

3 years ago

If Pablo invest $4,200 for 3 years and earn $630 what is the simple interest rate

Mariana [72]

The interest rate would be 5%

6 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:

<u><em>Question 1 a. i) Find the value of x.</em></u>

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

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:

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

<u><em>Question 1a ii) Find the volume of the frustrum</em></u>

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

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

<u><em>Question 1b. Calculate the volume of the bin</em></u>

<u>i) Upper frustrum</u>

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

<u>ii) Lower frustrum</u>

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

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

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

$Volume=38.3m^3$

<u>iii) Add the volume of the two frustrums</u>

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

6 0

3 years ago

If Amy has two apples and drops one, how many does she have now?

AURORKA [14]

She still has two since she only dropped it and hasn’t eaten it yet

8 0

2 years ago

The solution to w/-7 = -21 is 3.<br><br> True<br> False

Nat2105 [25]

False because -21/-7=w is 3 and 3/-7 cannot be -21

7 0

3 years ago

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