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Temka [501]
4 years ago
13

|x - 3| +8=3 Absolute value equations

Mathematics
2 answers:
MAXImum [283]4 years ago
6 0

|x - 3| = 3 - 8

|x -3| = -5

x - 3 = -5

x = -2

x-3 = 5

x = 8

x = 8, -2

But if you try and check that these solutions work, you will find that they do not. Since the absolute value equation has a negative value, there is no real solution.

kherson [118]4 years ago
5 0

Answer: No Solution

Step-by-step explanation:

|x-3| + 8 = 3

subtract 8 from both sides

|x-3| = -5

There is no real solution since <em>absolute value equation cannot be negative</em>

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In parallelogram HIJK, the measure of angle H is 45 degrees. 1.Find the measure of J and explain how you know
s2008m [1.1K]

Answer:

45°

Step-by-step explanation:

A parallelogram is a polygon with 4 sides and 4 angles (that is a quadrilateral).

For a parallelogram ABCD, Opposite sides and opposite angles are equal (i.e. BC = AD and ∠A = ∠C).

Consecutive angles are supplementary that is ∠A + ∠B = 180°.

The diagonals of a parallelogram bisect each other also separating the parallelogram into two congruent triangles.

In parallelogram HIJK, ∠H and ∠J are opposite angles. hence:

∠H = ∠J (opposite angles of a parallelogram are equal)

∠J = 45°

8 0
3 years ago
Richard purchased some books for $9.25 each. He paid a total of $46.25. If b is the number of books, which equation can be used
tamaranim1 [39]

Answer:

9.25b=46.25

Step-by-step explanation:

9.25b=46.25/:9.25

b=5

Richard purchased 5 books for $9.25 each

7 0
3 years ago
The ratio of 6 inches to 3 feet expressed in simplest form is 1/6. True False
erastovalidia [21]
The answer is true. Hope this helps!!!
3 0
3 years ago
Read 2 more answers
Find a particular solution to the nonhomogeneous differential equation y′′+4y=cos(2x)+sin(2x).
I am Lyosha [343]
Take the homogeneous part and find the roots to the characteristic equation:

y''+4y=0\implies r^2+4=0\implies r=\pm2i

This means the characteristic solution is y_c=C_1\cos2x+C_2\sin2x.

Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form y_p=ax\cos2x+bx\sin2x. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.

With y_1=\cos2x and y_2=\sin2x, you're looking for a particular solution of the form y_p=u_1y_1+u_2y_2. The functions u_i satisfy

u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx

where W(y_1,y_2) is the Wronskian determinant of the two characteristic solutions.

W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2

So you have

u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx
u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x

u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx
u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x

So you end up with a solution

u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x

but since \cos2x is already accounted for in the characteristic solution, the particular solution is then

y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x

so that the general solution is

y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x
7 0
3 years ago
19) *
RUDIKE [14]
The correct answer is 18 sides.
7 0
3 years ago
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