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3 years ago

Find the area of the trapezoid?

Mathematics

2 answers:

Verdich [7]3 years ago

8 0

Answer: 52.5 units squared

Step-by-step explanation:

The formula to find the area of a trapezoid is, (base1 + base2)/2 * height

So you find out which variable is which

Base 1 = 4

Base 2 = 3

Height = 15

Then plug it into the formula,

(4+3)/2 * 15

Then solve the order of operations

7/2 * 15

3.5 * 15

52.5 is the answer

KIM [24]3 years ago

8 0

Answer:

52.5

Step-by-step explanation:

Formula

Area = (1/2) (a + b)*h

Givens

a = 4

b = 3

h = 15

Solution

Area = (4 + 3)*15/2

Area = 7 * 15/2

Area = 105/2

Area = 52.5

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How many twenty-fourths would five sixths equal

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$n\cdot \frac { 1 }{ 24 } =\frac { 5 }{ 6 } \\ \\ 24\cdot n\cdot \frac { 1 }{ 24 } =\frac { 5 }{ 6 } \cdot 24$

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3 years ago

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7 Raja is laying tiles on a path that forms

ser-zykov [4K]

Answer:

The perimeter of the garden is 400 feet.

Step-by-step explanation:

Consider the provided information,

Raja is laying tiles on a path that forms the diagonal of a square garden. If Raja is told that the length of the diagonal path is $100\sqrt{2}$

The length of the diagonal of a square garden is $100\sqrt{2}$

Let x represent the side of square.

By Pythagorean theorem we know: $a^2+b^2=c^2$

Where c is hypotenuse.

The diagonal of the square divides the square into two right angle triangle.

Where, the sides of the square are the legs of the triangle and diagonal is the hypotenuse of the triangle.

$x^2+x^2=(100\sqrt{2})^2$

$2x^2=(100\sqrt{2})^2$

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Hence, the side of the square is 100 ft.

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3 years ago

Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficie

Dvinal [7]

For part (a), you have

$\dfrac x{x^2+x-6}=\dfrac x{(x+3)(x-2)}=\dfrac a{x+3}+\dfrac b{x-2}$
$x=a(x-2)+b(x+3)$

If $x=2$ , then $2=b(2-3)\implies b=-2$ .

If $x=-3$ , then $-3=a(-3-2)\implies a=\dfrac35$ .

So,

$\dfrac x{x^2+x-6}=\dfrac 3{5(x+3)}-\dfrac 2{x-2}$

For part (b), since the degrees of the numerator and denominator are the same, you first need to find the quotient and remainder upon division.

$\dfrac{x^2}{x^2+x+2}=\dfrac{x^2+x+2-x-2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}$

In the remainder term, the denominator $x^2+x+2$ can't be factorized into linear components with real coefficients, since the discriminant is negative $(1-4\times1\times2=-7)$ . However, you can still factorized over the complex numbers, so a partial fraction decomposition in terms of complexes does exist.

$x^2+x+2=0\implies x=-\dfrac12\pm\dfrac{\sqrt7}2i$
$\implies x^2+x+2=\left(x-\left(-\dfrac12+\dfrac{\sqrt7}2i\right)\right)\left(x-\left(-\dfrac12-\dfrac{\sqrt7}2i\right)\right)$
$\implies x^2+x+2=\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)$

Then you have

$\dfrac{x+2}{x^2+x+2}=\dfrac a{x+\dfrac12-\dfrac{\sqrt7}2i}+\dfrac b{x+\dfrac12+\dfrac{\sqrt7}2i}$
$x+2=a\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)+b\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)$

When $x=-\dfrac12-\dfrac{\sqrt7}2i$ , you have

$-\dfrac12-\dfrac{\sqrt7}2i+2=b\left(-\dfrac12-\dfrac{\sqrt7}2i+\dfrac12-\dfrac{\sqrt7}2i\right)$
$\dfrac32-\dfrac{\sqrt7}2i=-\sqrt7ib$
$b=\dfrac12+\dfrac3{2\sqrt7}i=\dfrac1{14}(7+3\sqrt7i)$

When $x=-\dfrac12+\dfrac{\sqrt7}2i$ , you have

$-\dfrac12+\dfrac{\sqrt7}2i+2=a\left(-\dfrac12+\dfrac{\sqrt7}2i+\dfrac12+\dfrac{\sqrt7}2i\right)$
$\dfrac32+\dfrac{\sqrt7}2i=\sqrt7ia$
$a=\dfrac12-\dfrac3{2\sqrt7}i=\dfrac1{14}(7-3\sqrt7i)$

So, you could write

$\dfrac{x^2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}=1-\dfrac {7-3\sqrt7i}{14\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)}-\dfrac {7+3\sqrt7i}{14\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)}$

but that may or may not be considered acceptable by that webpage.

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3 years ago

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