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2 years ago

Can someone please help me

Mathematics

2 answers:

pav-90 [236]2 years ago

8 0

Answer:

Step-by-step explanation:

First we get the same denominator by finding a common multiple.

20 is the lowest common multiple so let's use that.

9/10 will have to change so that the denominator is 20.

From 10 to 20 we have multiplemultiplied it by 2 so we will do the same to the numerator.

9 × 2 = 18.

The fraction is now 18/20.

Now 3/4 – we will get to 20 by multiplying by 5.

3 × 5 = 15

Now we can compare

18/20 and 15/20

9/10 and 3/4

MrRa [10]2 years ago

6 0

Hi sry I’m late but the answer is 18/20 and 15/20 and the other one is 9/10 and 3/4!!

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A rock thrown vertically upward from the surface of the moon at a velocity of 36m/sec reaches a height of s = 36t - 0.8 t^2 met

Verdich [7]

Answer:

a. The rock's velocity is $v(t)=36-1.6t \:{(m/s)}$ and the acceleration is $a(t)=-1.6 \:{(m/s^2)}$

b. It takes 22.5 seconds to reach the highest point.

c. The rock goes up to 405 m.

d. It reach half its maximum height when time is 6.59 s or 38.41 s.

e. The rock is aloft for 45 seconds.

Step-by-step explanation:

Velocity is defined as the rate of change of position or the rate of displacement. $v(t)=\frac{ds}{dt}$
Acceleration is defined as the rate of change of velocity. $a(t)=\frac{dv}{dt}$

The rock's velocity is the derivative of the height function $s(t) = 36t - 0.8 t^2$

$v(t)=\frac{d}{dt}(36t - 0.8 t^2) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(36t\right)-\frac{d}{dt}\left(0.8t^2\right)\\\\v(t)=36-1.6t$

The rock's acceleration is the derivative of the velocity function $v(t)=36-1.6t$

$a(t)=\frac{d}{dt}(36-1.6t)\\\\a(t)=-1.6$

b. The rock will reach its highest point when the velocity becomes zero.

$v(t)=36-1.6t=0\\36\cdot \:10-1.6t\cdot \:10=0\cdot \:10\\360-16t=0\\360-16t-360=0-360\\-16t=-360\\t=\frac{45}{2}=22.5$

It takes 22.5 seconds to reach the highest point.

c. The rock reach its highest point when t = 22.5 s

Thus

$s(22.5) = 36(22.5) - 0.8 (22.5)^2\\s(22.5) =405$

So the rock goes up to 405 m.

d. The maximum height is 405 m. So the half of its maximum height = $\frac{405}{2} =202.5 \:m$

To find the time it reach half its maximum height, we need to solve

$36t - 0.8 t^2=202.5\\36t\cdot \:10-0.8t^2\cdot \:10=202.5\cdot \:10\\360t-8t^2=2025\\360t-8t^2-2025=2025-2025\\-8t^2+360t-2025=0$

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=-8,\:b=360,\:c=-2025:\\\\t=\frac{-360+\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2-\sqrt{2}\right)}{4}\approx 6.59\\\\t=\frac{-360-\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2+\sqrt{2}\right)}{4}\approx 38.41$

It reach half its maximum height when time is 6.59 s or 38.41 s.

e. It is aloft until s(t) = 0 again

$36t - 0.8 t^2=0\\\\\mathrm{Factor\:}36t-0.8t^2\rightarrow -t\left(0.8t-36\right)\\\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\\\t=0,\:t=45$

The rock is aloft for 45 seconds.

5 0

4 years ago

gregori [183]

It’s D bc Dt= 15x3 and it’s not that hard to solve

6 0

4 years ago

What is the surface area of the triangular pyramid shown? Round to the nearest tenth.

Sloan [31]

Answer:

A. 57.6

Step-by-step explanation:

6 times 5.2 is 31.2. 4 times 5 is 20. 6 times 5 is 30. 6 times 5 is 30. (31.2/2)+(20/2)+30= 55.6 Then you round nearest tenth. 57.6.

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3 years ago

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gizmo_the_mogwai [7]

Attach a pic please
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3 years ago

At an university 70% of the students who attend stay on campus. If 1,380 of the students who attend live off campus, what is the

Alborosie

Answer:

4600

Step-by-step explanation:

We can write a proportion to find the total amount who attend university using the information given. A proportion is two equivalent ratios set equal to each other. Since 70% live on campus, then 30% live off campus and we are told that number is 1,380.

$\frac{30}{100}=\frac{1380}{y}$

We will cross multiply the numerator of one ratio with denominator of the other. And then solve for y.

30y=100(1380)

30y=138000

y=4600.

There are 4600 students who attend the university.

7 0

3 years ago