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jek_recluse [69]
3 years ago
12

-x<-x+7(x-2) include all of the steps to how you found your answer!

Mathematics
1 answer:
Lady bird [3.3K]3 years ago
6 0
-x<-x+7(x-2)
-x<-x+7x-14 Distribute the 7
-x<6x-14 Combine like terms
-7x<-14 Move all variables to one side
x>2 Divide by -7 to isolate the variable
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Can eat 10 cookies every 4 hours. How many hours would it take for her to eat 67 cookies?​
trapecia [35]

67 ÷ 10 x 4 = 26.8 hours for her to eat 67 cookies

6 0
3 years ago
What value of x is in the solution set of 3(x-4) &gt; 5x +2<br><br>-10<br>-5<br>5<br>10<br>​
ArbitrLikvidat [17]

Answer:

-10

Step-by-step explanation:

6 0
3 years ago
How do you find measure a and c
astra-53 [7]
You can use the measure of 112 on the similar shape to find b, as they are same side exterior angles. From there, you can find both a and c, because a/b and b/c are supplementary angles. 

a = 68 degrees

c = 68 degrees
5 0
3 years ago
Read 2 more answers
What is true of a sequence of transformations that rotates an image and then translates it in order to map it onto another image
joja [24]

The true statement about the sequence of transformations is it includes exactly two rigid transformations.

<h3>How to determine the true statement?</h3>

The transformation statement is given as:

a sequence of transformations that rotates an image and then translates it in order to map it onto another image

This can be split as follows:

  • A sequence of transformations that rotates an image
  • Then translates it in order to map it onto another image

Translation and rotation are rigid transformations

This means that the size and the angle of the shape that is transformed will remain the same

Hence, the true statement about the sequence of transformations is it includes exactly two rigid transformations.

Read more about transformation at

brainly.com/question/4289712

#SPJ1

4 0
2 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
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