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3 years ago

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A pool company is creating an image for a client that has the look of a 3-D image for a family pool and a similar dog pool. Find

the value of b in the drawing below:
Quadrilaterals STUV and WXYZ are shown. Side ST measures 22, side TU measures 16, side UV measures 14, side VS measures a, side

1 answer:

Natasha_Volkova [10]3 years ago

3 0

Answer:

Step-by-step explanation:

As the two pools, Family pool and Dog pool, are similar, their interior angles are equal and their corresponding sides are in the same proportion:

LMNO ~ PQRS

so LM/PQ = MN/QR = NO/RS = OL/SP

substituting in the given values

4/2 = 6/c = a/5 = 7/b

c = 6/2 = 3

a = 5*2 = 10

b = 7/2 = 3.5Answer:

Step-by-step explanation:

As the two pools, Family pool and Dog pool, are similar, their interior angles are equal and their corresponding sides are in the same proportion:

LMNO ~ PQRS

so LM/PQ = MN/QR = NO/RS = OL/SP

substituting in the given values

4/2 = 6/c = a/5 = 7/b

c = 6/2 = 3

a = 5*2 = 10

b = 7/2 = 3.5

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2 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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3x-11=-1x-75<br> find the x value that would make this equation true

ki77a [65]

Answer:

x = -16

Step-by-step explanation:

3x-11 = x-75

bring x to other side and simplify

(4x = -64) ÷4

x = -16

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3 years ago

Find the surface area of the regular pyramid

love history [14]

Greetings from Brazil!

$4\dfrac{1}{2} =\dfrac{9}{2}$

$Af1=\dfrac{b\cdot h}{2} \cdot 4\\\\\\Af1=\dfrac{\dfrac{9}{2} \cdot 4}{2} \cdot 4\\\\\\Af1=\dfrac{\dfrac{36}{2} }{2} \cdot 4\\\\\\Af1=\dfrac{18}{2} \cdot 4\\\\Af1=9\cdot 4\\\\Af1=36~m^2$

$Af2=l^2\\\\Af2=(\dfrac{9}{2})^2\\\\\ Af2=\dfrac{81}{4}\\\\Af2=20,25~m^2$

$At=Af1+Af2\\At=36+20,25\\At=56,25~m^2$

I hope I helped you! =)

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3 years ago

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cestrela7 [59]

Mean = 74.4

Median = 83.5

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3 years ago