All categories

2 years ago

Simplify 6 cm : 1.5 m

Mathematics

1 answer:

Zolol [24]2 years ago

4 0

6cm=0.06m

0.06m : 1.5m

6m : 150m

1m : 35m

You might be interested in

Solve for the variable

liraira [26]

Answer:

y = 25

Step-by-step explanation:

$30=\frac{6}{5} y\\\\\frac{6}{5} y=30\\\\(\frac{6}{5} y)5=30*5\\\\6y=150\\\\\frac{6y=150}{6}\\\\\boxed{y=25}$

Hope this helps!

8 0

3 years ago

Please help!!!! Tell me whether it’s a equation or not pleasr

goldfiish [28.3K]

This answer is a equation, my teacher told me this

3 0

2 years ago

Zoologists are studying two newly discovered species of insects in a previously unexplored section of rain forest. They estimate

salantis [7]

The population Pa of insect A after t years is given by the equation
Pa = 1.3(1-0.038)^t
while the population Pb of insect B after t years is
Pb = 2.1(1-0.046)^t

We equate the above expressions to find the number of years t it will take the two populations to be equal:
Pa = Pb
1.3(1-0.038)^t = 2.1(1-0.046)^t
1.3(0.962)^t = 2.1(0.954)^t
These are the equations that can be used to determine how long it will be before the populations of the two species are equal.

We can now solve for t:
(0.962)^t / (0.954)^t = 2.1/1.3
(0.962/0.954)^t = 2.1/1.3
After taking the log of both sides of our equation, number of years t is
t = log (2.1/1.3) / log (0.962/0.954)
t = 57 years
Therefore, it will take 57 years for the population of insect A to equal the population of insect B.

8 0

3 years ago

Read 2 more answers

An explosion causes debris to rise vertically with an initial speed of 120 feet per second. The formula h equals negative 16 t s

Novay_Z [31]

Answer:

The debris will be at a height of 56 ft when time is <u>0.5 s and 7 s.</u>

Step-by-step explanation:

Given:

Initial speed of debris is, $s=120\ ft/s$

The height 'h' of the debris above the ground is given as:

$h(t)=-16t^2+120t$

As per question, $h(t)=56\ ft$ . Therefore,

$56=-16t^2+120t$

Rewriting the above equation into a standard quadratic equation and solving for 't', we get:

$-16t^2+120t-56=0\\\textrm{Dividing by -8 throughout, we get}\\\frac{-16}{-8}t^2+\frac{120}{-8}t-\frac{56}{-8}=0\\2t^2-15t+7=0$

Using quadratic formula to solve for 't', we get:

$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\t=\frac{-(-15)\pm \sqrt{(-15)^2-4(2)(7)}}{2(2)}\\\\t=\frac{15\pm \sqrt{225-56}}{4}\\\\t=\frac{15\pm\sqrt{169}}{4}\\\\t=\frac{15\pm 13}{4}\\\\t=\frac{15-13}{4}\ or\ t=\frac{15+13}{4}\\\\t=\frac{2}{4}\ or\ t=\frac{28}{4}\\\\t=0.5\ s\ or\ t=7\ s$

Therefore, the debris will reach a height of 56 ft twice.

When time $t=0.5\ s$ during the upward journey, the debris is at height of 56 ft.

Again after reaching maximum height, the debris falls back and at $t=7\ s$ , the height is 56 ft.

5 0

3 years ago

Find to the nearest degree, the measure of the smaller acute angle of a right triangle whose sides are 7, 24,

Semmy [17]

The smallest angle would be 16.26°, with the other angles being 90° & 73.74°

7 0

3 years ago