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3 years ago

Wah wah wah wah wah? (what is the GCF of 100 and 200?)

Mathematics

2 answers:

Usimov [2.4K]3 years ago

4 0

Answer:

The GCF of 100 and 200 is 100.

Vedmedyk [2.9K]3 years ago

3 0

Answer:

100 :)

Step-by-step explanation:

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One kilometer is equal to 0.62 miles. Which best describes the relationship?

Alex_Xolod [135]

C. non linear this will be the answer

5 0

3 years ago

Find the mode of 3, 6, 4, 3, 2, 4, 7, 8, 6, 3, 9

lions [1.4K]

Answer:

3 is the mode because it appears most often.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

When y=7.5 x=2 what is the value of y when x is 2 1/4

swat32

<h2>Answer:</h2>

The value of y is 8.4375

<h2>Step-by-step explanation:</h2><h3>Known :</h3>

y = 7.5
x = 2

<h3>Asked :</h3>

The value of y when x is 2.25

<h3>Solution :</h3>

2/7.5 = 2.25/y

Do a cross multiplication,

2/7.5 = 2.25/y

=> 20/75 = 2.25/y

=> 75 . 2.25 = 20y

=> 168.75 = 20y

Reverse the equation,

168.75 = 20y

=> 20y = 168.75

Find the value of y,

20y = 168.75

=> y = 8.4375

<h3>Conclusion :</h3>

y = 8.4375

7 0

3 years ago

A rock is thrown up from the ground at a velocity of 84 feet per second. The formula h = -16t2 + 84t gives the rock's height in

kenny6666 [7]

Answer:

h = 110.24 m

Step-by-step explanation:

It is given that,

A rock is thrown up from the ground at a velocity of 84 feet per second. The formula $h = -16t^2 + 84t$ gives the rock's height in feet after t second.

We need to find the maximum height of the rock. For maximum height put $\dfrac{dh}{dt}=0$

So,

$\dfrac{d( -16t^2 + 84t)}{dt}=0\\\\-32t+84=0\\\\t=\dfrac{84}{32}\\\\t=2.62\ s$

Put t = 2.62 s in the equation of height. So,

$h = -16(2.62)^2 + 84(2.62)\\\\h=110.24\ m$

So, the maximum height of the rock is 110.24 m.

7 0

3 years ago

In ΔOPQ, the measure of ∠Q=90°, the measure of ∠O=26°, and QO = 4.9 feet. Find the length of PQ to the nearest tenth of a foot.

Step2247 [10]

Given:

In ΔOPQ, m∠Q=90°, m∠O=26°, and QO = 4.9 feet.

To find:

The measure of side PQ.

Solution:

In ΔOPQ,

$m\angle O+m\angle P+m\angle Q=180^\circ$ [Angle sum property]

$26^\circ+m\angle P+90^\circ=180^\circ$

$m\angle P+116^\circ=180^\circ$

$m\angle P=180^\circ -116^\circ$

$m\angle P=64^\circ$

According to Law of Sines, we get

$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$

Using the Law of Sines, we get

$\dfrac{p}{\sin P}=\dfrac{o}{\sin O}$

$\dfrac{QO}{\sin P}=\dfrac{PQ}{\sin O}$

Substituting the given values, we get

$\dfrac{4.9}{\sin (64^\circ)}=\dfrac{PQ}{\sin (26^\circ)}$

$\dfrac{4.9}{0.89879}=\dfrac{PQ}{0.43837}$

$\dfrac{4.9}{0.89879}\times 0.43837=PQ$

$2.38989=PQ$

Approximate the value to the nearest tenth of a foot.

$PQ\approx 2.4$

Therefore, the length of PQ is 2.4 ft.

4 0

3 years ago