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nignag [31]
3 years ago
5

Mandy used the input and output in this table to write ratios she concluded that because they are not all equivalent this is not

proportional is she correct
Mathematics
2 answers:
Alexandra [31]3 years ago
5 0

Answer:

They Would be if they Have a pattern from left to right such as +2 or -5.

Step-by-step explanation:

If an example is counting up from 15 to 31 and 0 to 8 the pattern is Adding Two. So if the numbers decrease or increase in a certain way, then you can find out what it is and thn bam, your answer to the question.

34kurt3 years ago
4 0

Answer:

Sample Response: No. The last ratio is not written with the values in the same position as the others. It should be 50

10

to be consistent. If the ratio were written this way, then the ratios would all be equivalent . The relationship is proportional.

Step-by-step explanation:

edge 2020 sample response

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Simplify the radical expression. Please explain how you got your answer.<br><br> √25/4
astra-53 [7]
√25/4=5/2 because 5/2 times 5/2 equal 25/4. Hope it help!
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3 years ago
What value or values of x satisfy 2x + 5.31 = 4.2?
Softa [21]

Answer:  x can equal -3.2 or -7.4.

Step-by-step explanation:

The correct expression is:

2|x+5.3|=4.2

So, we have an absolute value equation, where it can be positive or negative:

  • Positive

2 (x+5.3)=4.2

2x+10.6 =4.2

2x=4.2-10.6

2x = -6.4

x= -6.4/2

x = -3.2

  • Negative

2[-(x+5.3)]=4.2

-2x-10.6 =4.2

-2x= 4.2+10.6

-2x= 14.8

x= 14.8/-2

x= -7.4

x can equal -3.2 or -7.4.

Feel free to ask for more if needed or if you did not understand something.  

3 0
3 years ago
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3 years ago
Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y
irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

y'(x)= x^2y(x)-\frac12y^2(x)

Putting the value of y'(x) in equation (1)

y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))

Substituting x =0 and h= 0.2

y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]

\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]    [∵ y(0) =3 ]

\Rightarrow y(0.2)\approx 2.7

Substituting x =0.2 and h= 0.2

y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]

\Rightarrow y(0.4)\approx  2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]

\Rightarrow y(0.4)\approx 1.9926

Substituting x =0.4 and h= 0.2

y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]

\Rightarrow y(0.6)\approx  1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]

\Rightarrow y(0.6)\approx 1.6593

Substituting x =0.6 and h= 0.2

y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]

\Rightarrow y(0.8)\approx  1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]

\Rightarrow y(0.6)\approx 0.8800

Substituting x =0.8 and h= 0.2

y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]

\Rightarrow y(1.0)\approx  0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]

\Rightarrow y(1.0)\approx 0.9152

Therefore the value of y(1)= 0.9152.

4 0
3 years ago
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