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3 years ago

Can someone please help me out? This is literally torture.

Mathematics

2 answers:

Shkiper50 [21]3 years ago

8 0

4 is 2:6 7 is 25 8 is 8

kupik [55]3 years ago

5 0

3. There are 4 friends. A quarter of 4 is 25%. You can prove this by multipling 25% by 4 of the friends to get 100%. There is a 25% chance of Jonathon arriving first and a 25% chance that Michel will arrive second.

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The period T of a pendulum in seconds depends on its length L in feet, and is given by the formula T=

Damm [24]

This is a matter of plugging in. Specifically 7 for L.

T = 2πsqrt(7/32)
T = (6.28319...)(0.46771...)
T is about 2.93871 seconds

4 0

3 years ago

Can someone help me on this

nydimaria [60]

Answer:

10√3 feet

Step-by-step explanation:

In a 30-60-90 triangle;

-Long leg = long leg

-Short leg = 1/2 * long leg

-2nd longest leg = short leg * √3

-Long leg = 20

-Short leg = 1/2 * long leg = 10

-2nd longest leg = short leg * √3 = 10√3

8 0

3 years ago

Read 2 more answers

Show me quadrilaterals shapes

Volgvan

Anything with 4 sides is a quadrilateral like a rhombus square trapezoid and diamond

8 0

4 years ago

Read 2 more answers

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

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

Thus, the required solution is

$y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.$

6 0

3 years ago

The graph of a system of linear equations is shown below. What is the solution of the system?

Iteru [2.4K]

Answer: (2,2)

Step-by-step explanation:

It is where the two lines intersect.

7 0

3 years ago