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2 years ago

13

Please help it’s my last question. I don’t understand much of Algebra I. I will give brainiest for who helps me within bout 10 m

inutes

1 answer:

Elena L [17]2 years ago

6 0

There are four monomials, abcd; e; fg and h2. So that means there are four terms.

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What is the equation of a line that passes through the point (5, −4) and is parallel to the line whose equation is 2x + 5y = 10?

3241004551 [841]

Rewrite the given equation in slope intercept form:

5 y = 10 – 2 x

y = - 0.4 x + 2

Since the two lines are parallel, therefore they have similar slope which is – 0.4, so the equation is:

y = - 0.4 x + b

Given that x = 5, y = -4, find for b:

-4 = - 0.4 (5) + b

b = -2

So the answer is:

y = - 0.4 x – 2

or

<span>2 x + 5 y = - 10</span>

7 0

3 years ago

Solve the differential equation by variation of parameters.<br><br> y''+y=secx

suter [353]

Answer:y=

Acosx+Bsinx +cosx ln(cosx)+x sinx

Step-by-step explanation:

given equation y''+y=secx

auxiliary equation

$p^2+1=0\\p=\pm i$

so CF is y=Acosx+Bsinx

now

$y_1(x)=cosx \ \ \ \ y_2(x)=sinx\\{y_1}'(x)=-sinx \ \ \ \ {y_2}'(x)=cosx$

using wronskian formula

$W=\begin{vmatrix}cosx &sinx \\ -sinx & cosx\end{vmatrix}$

= $cos^2x+sin^2x=1$

now f(x)=secx

$u=-\int \frac{f(x)y_2(x)}{W(x)}dx \ and \ v=\int \frac{f(x)y_1(x)}{W(x)}dx$

$u=-\int \frac{secx\times sinx}{1}dx \ and \ v=\int \frac{secx\times cosx}{1}dx$

$u=-\int tanx dx \ and \ v=\int {1}dx$

$u=ln(cosx) \ \ \ and \ \ v=x$

now particular integrals are

PI=cosx ln(cosx)+x sinx

total solution

y= C.F+P.I

y=Acosx+Bsinx +cosx ln(cosx)+x sinx

8 0

3 years ago

1. The diagonal of a quadrilateral

zalisa [80]

Consider the length of diagonal is 8.5 cm instead of 8.5 m because length of perpendiculars are in cm.

Given:

Length of the diagonal of a quadrilateral = 8.5 cm

Lengths of the perpendiculars dropped on it from the remaining opposite vertices are 3.5 cm and 4.5 cm.

To find:

The area of the quadrilateral.

Solution:

Diagonal divides the quadrilateral in 2 triangles. If diagonal is the base of both triangles then the lengths of the perpendiculars dropped on it from the remaining opposite vertices are heights of those triangles.

According to the question,

Triangle 1 : Base = 8.5 cm and Height = 3.5 cm

Triangle 2 : Base = 8.5 cm and Height = 4.5 cm

Area of a triangle is

$Area=\dfrac{1}{2}\times base \times height$

Using this formula, we get

$Area(\Delta 1)=\dfrac{1}{2}\times 8.5\times 3.5$

$Area(\Delta 1)=14.875$

and

$Area(\Delta 2)=\dfrac{1}{2}\times 8.5\times 4.5$

$Area(\Delta 2)=19.125$

Now, area of the quadrilateral is

$Area=Area(\Delta 1)+Area(\Delta 2)$

$Area=14.875+19.125$

$Area=34$

Therefore, the area of the quadrilateral is 34 cm².

6 0

2 years ago

Abs(sin(x) - sin(y)) ≤ abs(x-y)

barxatty [35]

|sin(x)−sin(y)|=|2sin(x−y2)cos(x+y2)|≤|2sin(x−y2)|

3 0

3 years ago

Use the graph of the function. Determine Over what interval(s) the function is positive or negative.

san4es73 [151]

Answer:

I can't see the numbers well, but the graph is decreasing from (-inf, minimum) and increasing from (minimum, inf). The minimum is the vertex. Hope this helps!

Step-by-step explanation:

4 0

2 years ago