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Contact [7]
3 years ago
7

Please help me please!!!

Mathematics
1 answer:
spayn [35]3 years ago
5 0

Answer:

Its option number d

Step-by-step explanation:

Hope it helps u

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The diameter of a circle is 18m. Eugene claims that the circumstances of the circle is 113.04m. What is the circumference of the
pogonyaev

Answer:

The circumference is 56.52 m

Step-by-step explanation:

Eugene likely made the mistake of doubling the diameter thinking that it was the radius which gave the answer double the actual one. The equation for finding a circumference is C=\piD where d is equal to the diameter and c is the circumference. All you do is multiply 18 and 3.14 to get your answer.

4 0
3 years ago
What number is 5he same ratio to 64 as 5 is to 8
frosja888 [35]

Answer:

Answer is 24 and 40.

Step-by-step explanation:

I'm not quite sure if i'm wrong i'm terribly sorry :(

4 0
2 years ago
Z2 + (2j – 3)z + (5 – j) = 0
BARSIC [14]

Answer:

2z+(2j-3)z+(5-j)=0

2z +2jz-3z+5-j=0

2z-3z+2jz-j=-5

-z+2jz-j=-5

2jz-z-j=-5

I hope this help

4 0
3 years ago
Jose invested $5000 into a fund with an interest rate of 4.8%.
Minchanka [31]
$5000 × 0.048×2= $480

Answer: $480
3 0
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°

3 0
3 years ago
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