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3 years ago

Mia rolls a pair of standard six-sided dice. Find the chance that the two numbers add to 8.

Mathematics

1 answer:

larisa [96]3 years ago

7 0

Answer:

3/6 chance 1/2 chance

Step-by-step explanation:

Six sides on a dice:

1 and 2 and 3 and 4 and 5 and 6

one chance is 2 and 6 added together, another is 4 and 4, another is 5 and 3 so there are 3 chances in 3/6 which a 1/2

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A ramp leading into a building makes a 15o angle with the ground. The end of the ramp is 10 feet from the base of the building.

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Verdich [7]

1. Rewrite the expression in terms of logarithms:

$y=x^x=e^{\ln x^x}=e^{x\ln x}$

Then differentiate with the chain rule (I'll use prime notation to save space; that is, the derivative of y is denoted y' )

$y'=e^{x\ln x}(x\ln x)'=x^x(x\ln x)'$

$y'=x^x(x'\ln x+x(\ln x)')$

$y'=x^x\left(\ln x+\dfrac xx\right)$

$y'=x^x(\ln x+1)$

2. Chain rule:

$y=\ln(\csc(3x))$

$y'=\dfrac1{\csc(3x)}(\csc(3x))'$

$y'=\sin(3x)\left(-\cot^2(3x)(3x)'\right)$

$y'=-3\sin(3x)\cot^2(3x)$

Since $\cot x=\frac{\cos x}{\sin x}$ , we can cancel one factor of sine:

$y'=-3\dfrac{\cos^2(3x)}{\sin(3x)}=-3\cos(3x)\cot(3x)$

3. Chain rule:

$y=e^{e^{\sin x}}$

$y'=e^{e^{\sin x}}\left(e^{\sin x}\right)'$

$y'=e^{e^{\sin x}}e^{\sin x}(\sin x)'$

$y'=e^{e^{\sin x}+\sin x}\cos x$

4. If you're like me and don't remember the rule for differentiating logarithms of bases not equal to e, you can use the change-of-base formula first:

$\log_2x=\dfrac{\ln x}{\ln2}$

Then

$(\log_2x)'=\left(\dfrac{\ln x}{\ln 2}\right)'=\dfrac1{\ln 2}$

So we have

$y=\cos^2(\log_2x)$

$y'=2\cos(\log_2x)\left(\cos(\log_2x)\right)'$

$y'=2\cos(\log_2x)(-\sin(\log_2x))(\log_2x)'$

$y'=-\dfrac2{\ln2}\cos(\log_2x)\sin(\log_2x)$

and we can use the double angle identity and logarithm properties to condense this result:

$y'=-\dfrac1{\ln2}\sin(2\log_2x)=-\dfrac1{\ln2}\sin(\log_2x^2)$

5. Differentiate both sides:

$\left(x^2-y^2+\sin x\,e^y+\ln y\,x\right)'=0'$

$2x-2yy'+\cos x\,e^y+\sin x\,e^yy'+\dfrac{xy'}y+\ln y=0$

$-\left(2y-\sin x\,e^y-\dfrac xy\right)y'=-\left(2x+\cos x\,e^y+\ln y\right)$

$y'=\dfrac{2x+\cos x\,e^y\ln y}{2y-\sin x\,e^y-\frac xy}$

$y'=\dfrac{2xy+\cos x\,ye^y\ln y}{2y^2-\sin x\,ye^y-x}$

6. Same as with (5):

$\left(\sin(x^2+\tan y)+e^{x^3\sec y}+2x-y+2\right)'=0'$

$\cos(x^2+\tan y)(x^2+\tan y)'+e^{x^3\sec y}(x^3\sec y)'+2-y'=0$

$\cos(x^2+\tan y)(2x+\sec^2y y')+e^{x^3\sec y}(3x^2\sec y+x^3\sec y\tan y\,y')+2-y'=0$

$\left(\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^{x^3\sec y}-1\right)y'=-\left(2x\cos(x^2+\tan y)+3x^2\sec y\,e^{x^3\sec y}+2\right)$

$y'=-\dfrac{2x\cos(x^2+\tan y)+3x^2\sec y\,e^{x^3\sec y}+2}{\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^{x^3\sec y}-1}$

7. Looks like

$y=x^2-e^{2x}$

Compute the second derivative:

$y'=2x-2e^{2x}$

$y''=2-4e^{2x}$

Set this equal to 0 and solve for x :

$2-4e^{2x}=0$

$4e^{2x}=2$

$e^{2x}=\dfrac12$

$2x=\ln\dfrac12=-\ln2$

$x=-\dfrac{\ln2}2$

7 0

3 years ago

A right triangle is plotted in an x y plane. The vertices of the triangle are as follows: (0, 0); (x sub 1, y sub 1) in the seco

Nataly [62]

Based on the calculations, the area of this triangle is equal to: A. Area = 1/2 × [√(x₂² + y₂²) × √(x₁² + x₁²)].

<h3>How to calculate the area of a triangle?</h3>

Mathematically, the area of a triangle can be calculated by using this formula:

Area = 1/2 × b × h

Where:

b is the base area.

h is the height.

Next, we would find the distance between points A and B:

Distance = √[(x₂ - x₁)² + (y₂ - y₁)²]

AB = √[(x₂ - 0)² + (y₂ - 0)²]

AB = √(x₂² + y₂²)

Also, we would find the distance between points A and C:

Distance = √[(x₁ - x₂)² + (y₁ - y₂)²]

AC = √[(x₁ - 0)² + (y₁ - y0)²]

AC = √(x₁² + x₁²).

Now, we can find the area of this triangle:

Area = 1/2 × b × h

Area = 1/2 × AB × AC

Area = 1/2 × [√(x₂² + y₂²) × √(x₁² + x₁²)]

Read more on area of triangle here: brainly.com/question/2391510

#SPJ1

3 0

2 years ago