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olga55 [171]
3 years ago
6

Given quadrilateral RSTU, determine if each pair of sides (if any) are parallel and which are perpendicular for the coordinates

of the vertices.
R(1, -3), S(4, -1), T(2, 2), U(-4, -2)

Click on the graphic to complete the answer.
Mathematics
1 answer:
Mrac [35]3 years ago
3 0
Slope RS: 2/3
Slope ST: -3/2
Slope TU: 4/6 = 2/3
Slope UR: -1/5

Sides RS and TU have the same slope, so they are parallel.
The slope of side ST is the negative reciprocal of the slopes of sides RS and TU, so side ST is perpendicular to sides RS and TU.
Side UR is not parallel or perpendicular to any other side.

The quadrilateral is a trapezoid.
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A teenager who is 5 feet tall throws an object into the air. The quadratic function LaTeX: f\left(x\right)=-16x^2+64x+5f ( x ) =
tia_tia [17]

Answer:

At approximately x = 0.08 and x = 3.92.

Step-by-step explanation:

The height of the ball is modeled by the function:

f(x)=-16x^2+64x+5

Where f(x) is the height after x seconds.

We want to determine the time(s) when the ball is 10 feet in the air.

Therefore, we will set the function equal to 10 and solve for x:

10=-16x^2+64x+5

Subtracting 10 from both sides:

-16x^2+64x-5=0

For simplicity, divide both sides by -1:

16x^2-64x+5=0

We will use the quadratic formula. In this case a = 16, b = -64, and c = 5. Therefore:

\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Substitute:

\displaystyle x=\frac{-(-64)\pm\sqrt{(-64)^2-4(16)(5)}}{2(16)}

Evaluate:

\displaystyle x=\frac{64\pm\sqrt{3776}}{32}

Simplify the square root:

\sqrt{3776}=\sqrt{64\cdot 59}=8\sqrt{59}

Therefore:

\displaystyle x=\frac{64\pm8\sqrt{59}}{32}

Simplify:

\displaystyle x=\frac{8\pm\sqrt{59}}{4}

Approximate:

\displaystyle x=\frac{8+\sqrt{59}}{4}\approx 3.92\text{ and } x=\frac{8-\sqrt{59}}{4}\approx0.08

Therefore, the ball will reach a height of 10 feet at approximately x = 0.08 and x = 3.92.

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2 years ago
Simplify 4x-3(x-2). I'm stuck xD
alexira [117]

Answer:

Simplified = x + 6

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2 years ago
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Which drawing shows two lines that intersect at Point a ax, tab, two lines ab, straight x a.​
leonid [27]

The drawing which shows two lines that intersect at Point, A is as represented in Choice 1: Top-left corner.

Discussion:

In geometry, a line or straight line was introduced by ancient mathematicians to represent straight objects with negligible width and depth.

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Ultimately, The drawing which shows two lines that intersect at Point, A are as represented in Choice 1: Top-left corner.

Read more on intersection of lines:

brainly.com/question/9047265

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2 years ago
Consider the differential equation: xy′(x2+7)y=cos(x)+e3xy. Put the differential equation into the form: y′+p(x)y=g(x), determin
icang [17]

Answer:

Linear and non-homogeneous.

Step-by-step explanation:

We are given that

\frac{xy'}{(x^2+7)y}=cosx+\frac{e^{3x}}{y}

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We have also find type of differential equation.

y'=\frac{(x^2+7)y}{x}(cosx+\frac{e^{3x}}{y}}

y'=\frac{(x^2+7)cosx}{x}y+\frac{(x^2+7)e^{3x}}{x}

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It is linear differential equation because  this equation is of the form

y'+P(x)y=g(x)

Compare it with first order first degree linear differential equation

y'+P(x)y=g(x)

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\frac{dy}{dx}=\frac{(x^2+7)(ycosx+e^{3x})}{x}

Homogeneous equation

\frac{dy}{dx}=\frac{f(x,y)}{g(x,y)}

Degree of f and g are same.

f(x,y)=(x^2+7)(ycosx+e^{3x}),g(x,y)=x

Degree of f and g are not same .

Therefore, it is non- homogeneous .

Linear and non-homogeneous.

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zubka84 [21]
The answer is <span>2^7 = 128<span>

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