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Travka [436]
2 years ago
13

(conditional statement) If anyone knows this lesson, could you please help me with it? Tonight's the due date.​

Mathematics
1 answer:
lana66690 [7]2 years ago
3 0

Answer:

Hello lad, i am a professional and i seek out videos for every persons questions. I search far and wide, i hope to get an administrator role on this app, here is your answers! It explains everything!

Copy this link and go to it! It leads to a you-tube video with all the answers about this topic!

Step-by-step explanation:

Link: https://youtu.be/dQw4w9WgXcQ

Please mark BRAINLIEST! I hope to get a role in this app!

You might be interested in
(x^4)^2 <br> how do I simplify this expression?
castortr0y [4]

Answer:

x^8

Step-by-step explanation:

(x^4)^2

~Apply power rule [ (a^b)^c = a^bc ]

x^4(2)

~Simplify

x^8

Best of Luck!

6 0
2 years ago
Read 2 more answers
Students of Jesus and Mary were divided into 6 teams for a football game. if the school comprises of 3,900 students,how many stu
gtnhenbr [62]

Answer:

49 students/7 teams= 7 students per team

49/7=7

i hope this work for you

and sory if im wrang

3 0
2 years ago
SOMEONE PLEASE HELP this is worth 20 points
Ulleksa [173]

Answer:18432

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Can someone please help me with one and make sure to explain please it an Emergency. Also please answer it correctly.
ivann1987 [24]

Answer:

Step-by-step explanation:

Surface area of a cylinder:

S. A. = 2πrh + 2πr²

Volume of a cylinder:

V = πr²h

~~~~~~~~~~~

2(π)(2.5)(x) + 2(π)(2.5)² = π(2.5)²(x)

5x + 12.5 = 6.25x

1.25x = 12.5

x = 10

4 0
2 years ago
Write the formula for Newton's method and use the given initial approximation to compute the approximations x_1 and x_2. f(x) =
Harman [31]

Answer:

x_{n+1} = x_{n} - \frac{f(x_{n} )}{f^{'}(x_{n})}

x_{1} = -10

x_{2} = -3.95

Step-by-step explanation:

Generally, the Newton-Raphson method can be used to find the solutions to polynomial equations of different orders. The formula for the solution is:

x_{n+1} = x_{n} - \frac{f(x_{n} )}{f^{'}(x_{n})}

We are given that:

f(x) = x^{2} + 21; x_{0} = -21

f^{'} (x) = df(x)/dx = 2x

Therefore, using the formula for Newton-Raphson method to determine x_{1} and x_{2}

x_{1} = x_{0} - \frac{f(x_{0} )}{f^{'}(x_{0})}

f(x_{0}) = x_{0} ^{2} + 21 = (-21)^{2} + 21 = 462

f^{'}(x_{0}) = 2*(-21) = -42

Therefore:

x_{1} = -21 - \frac{462}{-42} = -21 + 11 = -10

Similarly,

x_{2} = x_{1} - \frac{f(x_{1} )}{f^{'}(x_{1})}

f(x_{1}) = (-10)^{2} + 21 = 100+21 = 121

f^{'}(x_{1}) = 2*(-10) = -20

Therefore:

x_{2} = -10 - \frac{121}{20} = -10+6.05 = -3.95

5 0
3 years ago
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