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2 years ago

A six-sided die is rolled.

Mathematics

2 answers:

insens350 [35]2 years ago

8 0

Answer:

a.) 1/3

Step-by-step explanation:

1/6 is a 2 is rolled

1/6 if a 4 is rolled

so 2/6 if either are rolled

which simplifies to 1/3

Vinil7 [7]2 years ago

8 0

Answer:

A. 1/3

Step-by-step explanation:

First take the number of sides on a standard die, which would be 6. Which means there’s 6 possible numbers that can be rolled. You want 2 or 4 which is 2 of the possible 6 outcomes. So there’s a 2/6 chance that you’ll receive one of those 2 outcomes. Finally simplify 2/6 to 1/3.

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If f(x) = 3^x + 10x and g(x) = 2x - 4, find (f + g)(x)

kow [346]

Answer:

The Answer is D

Step-by-step explanation:

All you do is add both equations 3^x+10x+2x-4

then you add the like varibles which is 10x+2x=12x

So, the answer is 3^x+12x-4

6 0

3 years ago

Read 2 more answers

The volume of a cylinder, V, in terms of radius, r,

Igoryamba

Answer:

See explanation

Step-by-step explanation:

Volume of a cylinder = πr²h

Solve for r

V = π * r² * h

r² = V/πh

Find the square root of both sides

r = √V/πh

If height = 6 meters and volume = 300 cubic meters. Find r

r = √V/πh

= √(300/3.14*6)

= √(300/18.84)

= √15.923566878980

r = 3.9904344223381

Approximately,

r = 4 meters

5 0

3 years ago

Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

$y=(-2+t)e^{-5t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.$

So, the general solution of the given equation is

$y(t)=(A+Bt)e^{-5t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.$

According to the given conditions, we have

$y(0)=-2\\\\\Rightarrow A=-2$

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$