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algol [13]
3 years ago
6

Which of the following is equal to 5.3?

Mathematics
2 answers:
soldier1979 [14.2K]3 years ago
8 0
[4.6+2×(5_2)]÷2=7

4.6+2=4.8
4.8×5=240
240_2=238
238÷2=116
NNADVOKAT [17]3 years ago
6 0

Answer:

[4.6+2×(5_2)]÷2=7

4.6+2=4.8

4.8×5=240

240_2=238

238÷2=116

Step-by-step explanation:

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From first principle find the derivative of logx​
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Hello!

\large\boxed{\frac{dy}{dx}logx = \frac{1}{xln10}  }

Recall that:

logx can be rewritten as:

log_{10}x

Use the equation for the derivative of a log expression:

\frac{dy}{dx}log_{a}u = \frac{1}{ulna} * u'

Substitute in the values in the expression:

logx = \frac{1}{xln10} * 1 = \frac{1}{xln10}

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Solving Exponential and Logarithmic Equations In Exercise, solve for x.<br> 4e2x - 3 - 5 = 0
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Answer:

Value of x will be 0.1515

Step-by-step explanation:

We have given the equation 4e^{2x}-3-5=0

We have top fond the value of x using exponent and logarithmic property

So 4e^{2x}-3-5=0

4e^{2x}-8=0

4e^{2x}=8

e^{2x}=2

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3 years ago
If you doubled the radius of a sphere, how would the volume change? A. The volume would double. B. The volume would triple. C. T
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7 0
3 years ago
You are constructing a cardboard box from a piece of cardboard with the dimensions 2 m by 4 m. You then cut equal-size squares f
postnew [5]

Answer:

Dimensions:

l =3,15 m

w=1,15 m

x= 0,42 m   (height)

V(max) = 1,52 m³

Step-by-step explanation:

The cardboard is L = 4 m   and   W = 2 m

Let call x the length of the square to cut in each corner, then, volume of open box is:

For the side L       is  L - 2*x               l =  4 - 2*x

For the side W      is  W - 2*x             w=  2 - 2*x

The height              is  x

Volume of the open box, as function of x is:

V(x) = ( 4 -2x) * ( 2 - 2x) *x    ⇒  V(x) = ( 8 - 8x -4x + 4x²) *x

V(x) = ( 8 -12x + 4x² )*x     V(x) = 8x - 12x² + 4x³

V(x) = 8x - 12x² + 4x³

Taking derivatives on both sides of the equation

V´(x) = 8 - 24x + 12x²

V´(x) = 0       8  - 24x + 12x² = 0     reordering    12x² - 24x + 8  = 0

or    3x² - 6x + 2 = 0

A second degree equation. Solving for x

x₁,₂  = ( 6 ± √36 - 24 ) /6

x₁,₂  = ( 6 ± 3.46) /6

x₁  =  6 + 3,46 /6     x₁  = 1.58  we dismiss such solution because 1,58 * 2 = 3,15 and is bigger than 2 one of the side of the cardboard

x₂  =( 6 - 3,46 ) / 6

x₂  = 0,42 m

Dimensions of the open box

l  = 4 - 2*x     l  = 4 - 0,85      l  =  3,15 m

w = 2 -2*x    w  = 2 - 0,85     w = 1,15 m

x = 0,42 m

V(max) =3,15*1,15*0,42

V(max) = 1,52 m³

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