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statuscvo [17]
2 years ago
6

Whats 234445x237=? fhrerjtiognmiovmenr

Mathematics
2 answers:
Naddika [18.5K]2 years ago
5 0

Answer:

234445×237=55,563,465

devlian [24]2 years ago
4 0

************************************************************************************************

Answer:

55,563,465

Step-by-step explanation:

Look below.

Have a great day! :)

Answered by: QueenOreo

************************************************************************************************

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Solve for the x in the inequality 6(2−6x)≤4−35x 6 ( 2 − 6 x ) ≤ 4 − 35 x .
dolphi86 [110]

Answer:

8 ≤ x

Step-by-step explanation:

6(2 − 6x) ≤ 4 − 35x

Expand bracket;

12 - 36x ≤ 4 − 35x

Add 36x to both sides to get;

12 ≤ 4 + x

Subtract 4 from both sides to get;

8 ≤ x

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1 year ago
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The prime number is 53 because the others have factors other than itself and 1 but 53 doesn't
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What is the vertex form of 2x^2+10x-5
Nataly_w [17]
<span><span>Two Solutions
1. x =(10-√140)/4=(5-√<span> 35 </span>)/2= -0.458</span><span> 

2. x =(10+√140)/4=(5+√<span> 35 </span>)/2= 5.458</span></span>
7 0
3 years ago
Examine the function f(x)=x+(3/x). Find the point on the curve at which the tangent lines pass through the point (1, 1).
vazorg [7]
Base on the function that you give and the data that are given. The point on the curve at which the tangent lines pass through the point (1,1). Base on my calculation and through my analyzations i came up with an answer of <span>-2x+3 = x+3/x</span>
6 0
3 years ago
Find parametric equations for the path of a particle that moves along the circle x2 + (y − 1)2 = 16 in the manner described. (En
ArbitrLikvidat [17]

Answer:

a) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t, b) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t, c) x = 4\cdot \cos \left(t+\frac{\pi}{2}  \right), y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right).

Step-by-step explanation:

The equation of the circle is:

x^{2} + (y-1)^{2} = 16

After some algebraic and trigonometric handling:

\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = 1

\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = \cos^{2} t + \sin^{2} t

Where:

\frac{x}{4} = \cos t

\frac{y-1}{4} = \sin t

Finally,

x = 4\cdot \cos t

y = 1 + 4\cdot \sin t

a) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t.

b) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t.

c) x = 4\cdot \cos t'', y = 1 + 4\cdot \sin t''

Where:

4\cdot \cos t' = 0

1 + 4\cdot \sin t' = 5

The solution is t' = \frac{\pi}{2}

The parametric equations are:

x = 4\cdot \cos \left(t+\frac{\pi}{2}  \right)

y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)

7 0
3 years ago
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