3(2x+1) =-5-5+3x solve the problem

Would shadows be 3 dimensional in 4 dimenional space

azamat

4 Dimensional shapes can cast 3 dimensional shadows. ... Maybe a shadow is the fourth dimension, its been here all the time.

4 0

3 years ago

There are a total of 84 green cars, 35 blue cars, and 42 red cars parked in the parking lot. How many cars are parked in the par

Cloud [144]

Ok so you just have to add all of the quantities together so 84+35=119 and then 119+42=161 so there are 161 cars in the parking lot. Hope this helps

8 0

3 years ago

Read 2 more answers

the distance between two towns on a map is 6cm. the actual distance between 2 towns is 48km what is the scale on the map

kodGreya [7K]

Answer:

1 cm=8 km

Step-by-step explanation:

We can use a ratio to determine the scale. Take the distance on the map on the left and the "real distance" on the right

6 cm=48 km

Divide by 6

6 cm/ 6 = 48 km /6

1 cm=8 km

4 0

2 years ago

A committee consisting of 3 faculty members and 5 students is to be formed. Every committee position has the same duties and vot

Lemur [1.5K]

Answer:

In 16170 ways the committee can be formed.

Step-by-step explanation:

3 faculty members and 5 students are required to form a committee.

The eligible to serve on the committee are 7 faculty members and 11 students.

If each committee position has the same duties and voting rights.

Then, the number of ways of selecting 3 faculty members out of 7 eligible faculty members is $^7C_{3} = 35$ .

Again, the number of ways of selecting 5 students out of 11 eligible students is given by $^{11}C_{5} = 462$ .

Therefore, in (35 × 462) = 16170 ways the committee can be formed. (Answer)

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

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

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

3(2x+1) =-5-5+3x solve the problem ​

3(2x+1) =-5-5+3x solve the problem