Write an equation of the line containing the point (3,1) and perpendicular to the line 5x-3y=5

torisob [31]

-16 = 8x/5......8x/5 is the same as 8/5(x)

-16 = (8/5)x...divide both sides by 8/5
-16 / (8/5) = x
-16 * 5/8 = x
-80/8 = x
-10 = x <===

6 0

3 years ago

Stephanie has 60 tickets to the carnival.She used 3/4 of her 60 tickets to play games.Then she used 1/5 of her remaining tickets

Lera25 [3.4K]

Answer:

<h2>She gave her brother 6 tickets</h2>

Step-by-step explanation:

This problem is on fractional numbers.

given that she has 60 tickets

1.She used 3/4 of her tickets to play game

=(3/4)*60

=180/4

=45

She is left with (60-45)= 15 tickets

2. She used 1/5 of her remaining tickets for rides

=(1/5)*15

=15/5

=3

She is left with (15-3)= 12 tickets

3. She gave half of her tickets to her brother.

= (1/2)*12

=12/2

= 6

She gave her brother 6 tickets

3 0

3 years ago

You bought a large container of fruit punch. The label on the container says there are

erastovalidia [21]

There are 8 fl ounces in 1 cup.
Divide 128 by 8.
16 cups

6 0

3 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

Which operation results in a binomial? (3y ^ 6 + 4) (9y ^ 12 - 12y ^ 6 + 16)

AleksAgata [21]

Step-by-step explanation:

Given the expression, $(3y ^ 6 + 4) (9y ^ {12} - 12y ^ 6 + 16)$ , we want to know which sign placed in between the two polynomial will produce a binomial.

Using a plus (+) sign to check:

$= (3y ^ 6 + 4) + (9y ^ {12} - 12y ^ 6 + 16)\\= (3y ^ 6 + 4) + (9y ^ {12} - 12y ^ 6 + 16)\\collect \ like \ terms\\= 9y^{12} +3y^6 - 12y^6 + 16+ 4\\= 9y^{12} - 9y^6 + 20\\$

From the gotten value, it can be seen that the resulting answer produces 3 terms, showing that it results in a trinomial instead of binomial.

Using a plus (-) sign to check:

$= (3y ^ 6 + 4) - (9y ^ {12} - 12y ^ 6 + 16)\\= 3y ^ 6 + 4 - 9y ^ {12} + 12y ^ 6 - 16\\collect \ like \ terms\\= -9y^{12} +3y^6 +12y^6 + 4-16\\= -9y^{12} +15y^6 -12\\$

From the gotten value, it can be seen that the resulting answer produces 3 terms, showing that it results in a trinomial instead of binomial.

<em>Hence none of the operations given results in a binomial</em>

3 0

3 years ago