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2 years ago

GCF of 48, 36, and 16. (this is all I need thx) :)

Mathematics

1 answer:

erastova [34]2 years ago

5 0

Answer:

Step-by-step explanation:

16=2^4

48=2^4·3

36=2²·3²

GCF=2^2

hope this helps :)

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Suppose that weight of adolescents is being studied by a health organization and that the accompanying tables describes the prob

svet-max [94.6K]

Answer:

<em>hello your question lacks some information attached below is the complete question</em>

answer : NO

Step-by-step explanation:

Determine the probability of not selecting an obese subject in 3 trials

= P ( not obese in 3 randomly selected ) = ( 1 - 0.111 )^3

= 0.703

The probability > 0.05

hence it is insignificant therefore the answer is NO

6 0

2 years ago

In a triangle ABC, if <B = 60° and the ratio of two sides is a: c= 2: square root 3+1, then <A=

slavikrds [6]

Your answer is 75°.

To answer this question you need to use both the cosine rule and the sine rule. First, we need to find the length of side b by using the cosine rule, where a = 2 and c = √3 + 1. Then you substitute these into the equation:

b² = a² + c² - 2×a×c×cos(B)

b² = (2)² + (√3 + 1)² - 2×2×(√3 + 1)×cos(60)

b² = 4 + 4 + 2√3 - (4 - 4√3)×0.5

b² = 8 - 2 = 6

b = √6

Then you use this length in the sine rule, and find the angle:

$\frac{sin(A)}{\sqrt{3}+1 } =\frac{sin(60)}{\sqrt{6} }$

sin(A) = (√6 + √2)/4

A = 75

I hope this helps! Let me know if you have any questions

7 0

3 years ago

Help with 30 please. thanks.

Svet_ta [14]

Answer:

See Below.

Step-by-step explanation:

We have the equation:

$\displaystyle y = \left(3e^{2x}-4x+1\right)^{{}^1\! / \! {}_2}$

And we want to show that:

$\displaystyle y \frac{d^2y }{dx^2} + \left(\frac{dy}{dx}\right) ^2 = 6e^{2x}$

Instead of differentiating directly, we can first square both sides:

$\displaystyle y^2 = 3e^{2x} -4x + 1$

We can find the first derivative through implicit differentiation:

$\displaystyle 2y \frac{dy}{dx} = 6e^{2x} -4$

Hence:

$\displaystyle \frac{dy}{dx} = \frac{3e^{2x} -2}{y}$

And we can find the second derivative by using the quotient rule:

$\displaystyle \begin{aligned}\frac{d^2y}{dx^2} & = \frac{(3e^{2x}-2)'(y)-(3e^{2x}-2)(y)'}{(y)^2}\\ \\ &= \frac{6ye^{2x}-\left(3e^{2x}-2\right)\left(\dfrac{dy}{dx}\right)}{y^2} \\ \\ &=\frac{6ye^{2x} -\left(3e^{2x} -2\right)\left(\dfrac{3e^{2x}-2}{y}\right)}{y^2}\\ \\ &=\frac{6y^2e^{2x}-\left(3e^{2x}-2\right)^2}{y^3}\end{aligned}$