What is th vaule of log4 16 ?

Mathematics

1 answer:

FromTheMoon [43]3 years ago

5 0

Answer:

the vaule of log4 16 is 2

Step-by-step explanation:

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Y=4x(8-x). Linear or nonlinear? explain.

NemiM [27]

Answer:

not linear

Step-by-step explanation:

this equation becomes y = -4x² + 32x which is a quadratic equation that, when graphed, forms an upside down u-shaped curve (upside down because the x² term is negative; if it's positive then the curve is u-shaped)

4 0

2 years ago

Read 2 more answers

Unit 1: Scientific Notation Constructed Response Assignment

balu736 [363]

brainly.com/question/5027087

Let C, S and H be the diameters of cell, sand and hair

Diameter of a cell: $C=1 \cdot 10^{-5}m$

Diameter of a grain of sand: $S=2 \cdot10^{-4}m$

Part A:

$H=0.000025=2.5 \cdot 10^{-5}$ meters, because it takes 5 movements right of the decimal dot, until there is only one digit to its left.

Part B:

we are comparing $S=2 \cdot10^{-4} m$ to $C=1 \cdot10^{-5} m$ ,

$\frac{S}{C}= \frac{2 \cdot 10^{-4}}{1 \cdot 10^{-5}}=\frac{20 \cdot 10^{-5}}{1 \cdot 10^{-5}}=20$

thus S is 20 times larger than C

Part C

$1m=1 \cdot10^9nm$

thus,

$H=2.5 \cdot 10^{-5}m=2.5 \cdot 10^{-5} \cdot 1 \cdot 10^9nm=2.5 \cdot 10^{-5+9}=2.5\cdot10^4nm$

Part D.

$1m=1\cdot 10^9nm\\\\1nm=\displaystyle{ \frac{1m}{10^9}=1\cdot 10^{-9}m}$

thus 300 nm are written in meters as follows:

$300nm= 300\cdot 1\cdot 10^{-9}m=3\cdot 10^2 \cdot 10^{-9}m=3\cdot 10^{-7}m$

5 0

3 years ago

At what point does the curve have maximum curvature? y = 9 ln(x) (x, y) =

Andrews [41]

y = 9ln(x)
y' = 9x^-1 =9/x
y'' = -9x^-2 =-9/x^2

curvature k = |y''| / (1 + (y')^2)^(3/2)

= |-9/x^2| / (1 + (9/x)^2)^(3/2)
= (9/x^2) / (1 + 81/x^2)^(3/2)
= (9/x^2) / [(1/x^3) (x^2 + 81)^(3/2)]
= 9x(x^2 + 81)^(-3/2).

To maximize the curvature,

we find where k' = 0. 
k' = 9 * (x^2 + 81)^(-3/2) + 9x * -3x(x^2 + 81)^(-5/2)
...= 9(x^2 + 81)^(-5/2) [(x^2 + 81) - 3x^2]
...= 9(81 - 2x^2)/(x^2 + 81)^(5/2)

Setting k' = 0 yields x = ±9/√2.

Since k' < 0 for x < -9/√2 and k' > 0 for x > -9/√2 (and less than 9/√2),
we have a minimum at x = -9/√2.

Since k' > 0 for x < 9/√2 (and greater than 9/√2) and k' < 0 for x > 9/√2,
we have a maximum at x = 9/√2.

x=9/√2=6.36

y=9 ln(x)=9ln(6.36)=16.66

the answer is
(x,y)=(6.36,16.66)

7 0

3 years ago

1) Solve the following system of equations algebraically. Give your solution as two

Bad White [126]

For this case we have the following system of equations:

$y = x ^ 2 + 7x + 12\\y = 2x + 8$

Equating both equations we have:

$x ^ 2 + 7x + 12 = 2x + 8\\x ^ 2 + 7x-2x + 12-8 = 0\\x ^ 2 + 5x + 4 = 0$

We must find the solutions, for this we factor. We look for two numbers that, when multiplied, result in 4 and when added, result in 5. These numbers are 4 and 1:

$4 + 1 = 5\\4 * 1 = 4$