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4 years ago

The endpoints of a side of rectangle ABCD in the coordinate plane are at A (2, 11) and

Mathematics

1 answer:

Doss [256]4 years ago

5 0

ANSWER

$2y-x-20=0$

$y=\frac{1}{2}x+10$

EXPLANATION

Let the rectangle be oriented as shown in the diagram.

The line segment AD passes through $A(2,11)$ .

All we need now is the slope of AD then we can find its equation.

Since AD is perpendicular t AB, we determine the slope of AB and then use it to find the slope of AD.

$Slope_{AB}=\frac{1-11}{7-2}$

$Slope_{AB}=\frac{-10}{5}=-2$

The slope of AD is the negative reciprocal of the slope of AB because they are perpendicular.

$Slope_{AD}=\frac{-1}{-2}=\frac{1}{2}$

The equation of AD is given by;

$y-y_1=m(x-x_1)$

$y-11=\fra[1}{2}(x-2)$

Multiplying through by 2 gives,

$2y-22=(x-2)$

$2y-x-22+2=0$

$2y-x-20=0$

$y=\frac{1}{2}x+10$

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3 years ago

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Please help me 33/2 + 3y/5 = 7y/10 + 15

8090 [49]

We are given with an equation in variable y and we need to solve for y . So , now let's start !!!

We are given with ;

${:\implies \quad \sf \dfrac{33}{2}+\dfrac{3y}{5}=\dfrac{7y}{10}+15}$

Take LCM on both sides :

${:\implies \quad \sf \dfrac{165+6y}{10}=\dfrac{7y+150}{10}}$

Multiplying both sides by 10 ;

${:\implies \quad \sf \cancel{10}\times \dfrac{165+6y}{\cancel{10}}=\cancel{10}\times \dfrac{7y+150}{\cancel{10}}}$

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Can be further written as ;

${:\implies \quad \sf 7y+150=165+6y}$

Transposing 6y to LHS and 150 to RHS

${:\implies \quad \sf 7y-6y=165-150}$

${:\implies \quad \bf \therefore \quad \underline{\underline{y=15}}}$

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2 years ago

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3 years ago

Find the derivative.

Aleksandr [31]

Answer:

Using either method, we obtain: $t^\frac{3}{8}$

Step-by-step explanation:

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$\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du$

The integral itself can be evaluated by writing the root and exponent of the variable u as: $\sqrt[8]{u^3} =u^{\frac{3}{8}$

Then, an antiderivative of this is: $\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}$

which evaluated between the limits of integration gives:

$\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}$

and now the derivative of this expression with respect to "t" is:

$\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}$

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

$g(x)=\int\limits^x_a {f(t)} \, dt$

is continuous on [a,b], differentiable on (a,b) and $g'(x)=f(x)$

Since this this function $u^{\frac{3}{8}$ is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

$\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}$

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