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2 years ago

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The area of a rectangular pool is given by the expression, h^2+11h+28. Which of the following represents the length and width of

the pool, in terms of h?
A) (h+2)(h+14)

B) (h+1)(h+28)

C) (h+4)(h+28)

1 answer:

BaLLatris [955]2 years ago

6 0

None of the solutions above.

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What is a correct first step in solving the inequality –4(3 – 5x)≥ –6x + 9?

Maksim231197 [3]

Hello,
the first step should be to distribue:
-4(3-5x)= -12+20x

Teh resolution may be:

-4(3-5x)>=-6x+9
==>-12+20x>=-6x+9
==>20x+6x>=9+12
==>26x>21
==>x>=21/26

But an other way may be used:

-4(3-5x)>=-6x+9
==>3-5x<= -6x/(-4)+9/(-4)
==>-5x-3/2 x<=-9/4 -3

==>-13/2 x <=-21/4

==>x>= -21/4 *(-2/13)
==>x>=21/(2*13)
==>x>=21/26

5 0

3 years ago

Read 2 more answers

Read the question and type your response in the box provided. Your response will be saved automatically.

larisa [96]

Answer:

512

Step-by-step explanation:

8 x 8 x 8 = 512

4 0

2 years ago

Read 2 more answers

CA Geometry A Illuminate Credit 4 FF.pdf<br> all 4 pls

lions [1.4K]

Answer:

1) HL=LJ=63

(ZJ)²=16²+63²

(ZJ)²=256+3969

ZJ=√4425

ZJ= 65

2)HJ=HL+LJ

= 63+63

HJ=126

3) XC=XA=8

4) BE=AB/2=20/2=10

<u>OAmalOHopeO</u>

6 0

3 years ago

Please see attachment

Dafna11 [192]

Answer:

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

Step-by-step explanation:

<u>Step(i)</u>:-

Given function

$f(x) = \frac{-x}{2x^{2} +1}$ ...(i)

Differentiating equation (i) with respective to 'x'

$f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2} }$ ...(ii)

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} }$

Equating Zero

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$\frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$2 x^{2}-1 = 0$

$2 x^{2} = 1$

$x^{2} = \frac{1}{2}$

$x = \frac{-1}{\sqrt{2} } , x = \frac{1}{\sqrt{2} }$

<u><em>Step(ii):</em></u>-

Again Differentiating equation (ii) with respective to 'x'

$f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4} }$

put

$x = \frac{1}{\sqrt{2} }$

$f^{ll} (x) > 0$

The absolute minimum value at $x = \frac{1}{\sqrt{2} }$

<u><em>Step(iii):</em></u>-

The value of absolute minimum value

$f(x) = \frac{-x}{2x^{2} +1}$

$f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}$

on calculation we get

The value of absolute minimum value = - 0.3536

<u><em>Final answer</em></u>:-

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

3 0

2 years ago

Please help work this out

BigorU [14]

Answer:

Area=190.091 cm^2

Step-by-step explanation:

Area = 1/2(Pi x r^2) one-half because it's a semi-circle

Area=1/2(3.14 x 11^2)

11^2=121 so, Area=1/2(3.14 x 121)

Area=1/2(379.94)

Area=189.97 cm^2

adjustment:

Area=1/2(3.142 x 11^2)

Area=1/2(3.142 x 121)

Area=1/2(380.182)

Area=190.091

5 0

2 years ago