Ask question

Ask question

All categories

2 years ago

15

On Monday, Sam ran one mile. The distance he runs each day thereafter, is 15% greater than the distance Sam runs first exceeds t

wo miles is which day of the week?

1 answer:

poizon [28]2 years ago

7 0

There are some grammar mistakes and therefore I can't read it properly to answer

You might be interested in

Which of the following is a rational function

Anestetic [448]

B is the only rational answer.

3 0

3 years ago

Read 2 more answers

What is this fraction in simplest form 12/75 please help me this homework is so hard my teacher said just try your best

garri49 [273]

Answer:

4/25

Step-by-step explanation:

12 divided by 3

72 divided by 3

Greatest Common Divisor:

3

So 12/72 simplifed would be 4/25 and the GCD would be 3.

8 0

2 years ago

6x(x + 5) = 0<br> What this equal too

KatRina [158]

Answer:

x = -5

Step-by-step explanation:

-5 + 5 = 0 0×6=0

10100101010100101010101010101010101010

3 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

2 years ago

Can someone please help me ASAP

konstantin123 [22]

Answer:

x = 50

Step-by-step explanation:

Since MNP is a tangent to the circle at N, then

∠ ONP = 90° , so

∠ ONL = 90° - 55° = 35°

OL = ON ( both radii of the circle )

Thus Δ ONL is isosceles and the base angles are congruent

∠ OLN = ∠ ONL = 35°

JKLN is a cyclic quadrilateral whose opposite angles are supplementary, that is

∠KJN + ∠ KLN = 180 , substitute values

95 + 35 + x = 180

130 + x = 180 ( subtract 130 from both sides )

x = 50

4 0

2 years ago