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lbvjy [14]
4 years ago
9

Evaluate. Math problem 7th grade.

Mathematics
1 answer:
marshall27 [118]4 years ago
5 0

The answer is -12

please see the attached picture for full solution

hope it helps

Good luck on your assignment..

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Please help me i'll give 20 points to whoever helps me answer all 4 of them and brainly
Brilliant_brown [7]

Answer:

G - number of arrangements

F. the amount of her allowance

B. y = x

H. the number of hours

7 0
3 years ago
NEED HELP!!! please:)
Dahasolnce [82]
No, Jonah is not correct. The answer is shown in the picture.

8 0
3 years ago
Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu
Alisiya [41]
Let y=C_1x+C_2x^3=C_1y_1+C_2y_2. Then y_1 and y_2 are two fundamental, linearly independent solution that satisfy

f(x,y_1,{y_1}',{y_1}'')=0
f(x,y_2,{y_2}',{y_2}'')=0

Note that {y_1}'=1, so that x{y_1}'-y_1=0. Adding y'' doesn't change this, since {y_1}''=0.

So if we suppose

f(x,y,y',y'')=y''+xy'-y=0

then substituting y=y_2 would give

6x+x(3x^2)-x^3=6x+2x^3\neq0

To make sure everything cancels out, multiply the second degree term by -\dfrac{x^2}3, so that

f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y

Then if y=y_1+y_2, we get

-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0

as desired. So one possible ODE would be

-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0

(See "Euler-Cauchy equation" for more info)
6 0
3 years ago
PLEASE HELP!!! :))
Nina [5.8K]

For a better understanding of the solution given here please find the attached file which has the relevant diagram.

To answer this question we will have to make use of <u>the Isosceles Triangle Theorem</u> which states that the perpendicular bisector of the base of an isosceles triangle is also the angle bisector of the vertex angle. Thus, as a corollary we know that is EF is the angle bisector of the vertex angle ∠E, then, EF is the perpendicular bisector of the of the base DK.

Please follow the diagram of a complete understanding of the logic and the solution.

As EF is the angle bisector as given in the question, thus we will have:

m\angle DEK=2\times m\angle DE F=2\times 43^{\circ}=86^{\circ}.

Also, from the Theorem we know that KF will be half of DK and thus, KF will be:KF=\frac{1}{2}DK= \frac{1}{2}\times 16=8 centimeters.

Likewise, from the same theorem we have: m\angle EFD=90^{\circ}

4 0
3 years ago
PRETTY EASY PLS HELP I BEG U
lbvjy [14]

Answer:

Step-by-step explanation:

110 I think

3 0
4 years ago
Read 2 more answers
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