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3 years ago

(pls help .z.) Both the numerator and the denominator of 21/24 can be divided by _____. .2 .3 .4 .6

Mathematics

2 answers:

MAXImum [283]3 years ago

7 0

Answer:

Step-by-step explanation:

for us to find the answer we have to find the number that divides evenly to both numbers with no remainder.

2: 21/2= 10.5 you can stop here because you already got an uneven answer

3: 21/3= 7 and 24/3= 8. theres your answer! 3 divides into 21 and 24 evenly with no remainder

4: 21/4= 5.25 you can stop here because you already got an uneven answer

6: 21/6= 3.5 you can stop here because you already got an uneven answer

also because i got it right on oddysseyware :D

makkiz [27]3 years ago

3 0

Answer:

B.) 3

Step-by-step explanation:

21/3 = 7

24/3 = 8

Hopefully this helps you :)

pls mark brainlest ;)

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Answer:

12 years if interest is paid yearly

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Step-by-step explanation:

2.75% of $1500 is $41.25 interest per period

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3 years ago

The circle below has a radius of 3 units . Find the area of the circle. Use 3.14 for π.

const2013 [10]

Answer:

28.26 units

Step-by-step explanation:

We know the radius of the circle is 3 units, so we'll end up plugging that into the formula.

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3 years ago

-11 3(4-5a)=? -11 12-15a =1-15a

den301095 [7]

Answer = a=12/7=1.714

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3 years ago

Es 18 de junio de 1815, las tropas napoleónicas se encuentran justo adelante, tu eres el ingeniero en balística y el general Wel

Airida [17]

Answer:

89.1° or -1.4°

Step-by-step explanation:

1. Location:

You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.

The French troops are about 50 m below you and 1.2 km distant.

2. Finding the firing angle

Data:

R = 1200 m

u = 600 m/s

h = -50 m (the height of the target)

a = 9.8 m/s²

We have two conditions.

Horizontal distance

(1) 1200 = 600t cosθ

Vertical distance

(2) -50 = 600t sinθ - 4.9t²

Divide each side of (1) by 600cosθ.

$(3) \, t =\dfrac{2}{\cos \theta}$

Substitute (3) into (2)

$-50 = 600t \sin \theta - 4.9t^{2} = 600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}$

Recall that

(5) sec²θ = 1/cos²θ = tan²θ + 1

Substitute (5) into (4)

$-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )$

Set up a quadratic equation

$\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}$

Solve for θ

Use the quadratic formula.

tanθ = 61.249 or -0.025

θ = arctan(61.249) = 89.1° or

θ = arctan(-0.025) = -1.4°

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3 years ago

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bonufazy [111]

Answer:

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Step-by-step explanation:

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