Which expression is equivalent to 7(xy)??

Ms. Houston's fifth grade class is going to A museum the class raises $68 for the trip transportation to the museum cost $40The

damaskus [11]

68-40=28
28/4=7
therefore, 7 is your answer
(Hope it helps my friend ^_^)

8 0

3 years ago

Read 2 more answers

Answer the question 5+5+5+5+5

JulsSmile [24]

Answer:

25

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Number sense. What is the rule? <br> 200---->500 400---->700 500----->800

Dimas [21]

The rule would be addition. (+300)

6 0

3 years ago

Here is a balanced hanger diagram

suter [353]

Based on the mass of the circle and the triangle, we can find the mass of the square to be<u> 3.33 grams</u>

<h3>Mass of each side of hanger </h3>

Assuming the mass of the square is x, the equation for the first side is:

= (3 x mass of circle) + (2 x mass of triangle) + (6 x mass of square)

= ( 3 x 2) + ( 2 x 4) + ( 6 × x)

= 6 + 8 + 6x

Mass of other side:

= (2 x mass of circle) + (5 x mass of triangle) + (3 x mass of square)

= ( 2 x 2) + ( 5 x 4) + ( 3 × x)

= 4 + 20 + 3x

<h3>Mass of square </h3>

As both sides are equal, equate both formulas to find x:

6 + 8 + 6x = 4 + 20 + 3x

6x - 3x = 24 - 14

3x = 10

x = 10/3

x = 3.33 grams

In conclusion, each square is 3.33 grams.

Find out more on problems requiring equating at brainly.com/question/20213883.

8 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