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2 years ago

6

Solve for g:

2 answers:

Airida [17]2 years ago

7 0

Hey there!

Assuming you meant: $\frac{5}{8} = \frac{12}{g}$

If so, follow these steps! (if you want to!)

So, let's cross multiply

Like: $5(g) = 12(8)$

$5(g) = 5g \\ 12(8) = 96$

We get: $5g = 96$

Divide both sides of your equation by $5$

$\frac{5g}{5} = \frac{96}{5}$

Cancel out $\frac{5g}{5}$ because it'll give you a outcome of a $1$

Keep: $\frac{96}{5}$ because that one solve for your answer

$\frac{96}{5} = 19.2$

Answer to your equation: $19.2$

Good luck on your assignment and enjoy your day!

~ $LoveYourselfFirst:)$

yulyashka [42]2 years ago

5 0

$\dfrac{5}{8}=\dfrac{12}{g}\ \ \ |\text{cross multiply}\\\\5\cdot g=8\cdot12\\\\5g=96\ \ \ |:5\\\\g=19.2$

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The value of y varies directly with x. If x =3, then y=21. What is the value of x when y=105?

Goshia [24]

Answer:

x = 15

Step-by-step explanation:

given that y varies directly with x

mathematically we can express this as

y = k x , where k is a constant

step 1: find k

given x = 3 and y = 21

y = k x

21 = k (3)

k = 7

hence the equation becomes

y = 7 x

when y = 105,

105 = 7x

x = 105 / 7 = 15

3 0

2 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

An arch is in the shape of a parabola. It has a span of 100 feet and a maximum height of 25 ft.

faltersainse [42]

Answer:

$y=-\frac{1}{100}(x-50)^2+25$

the height of the arch 10 feet from the center is 24 feet

Step-by-step explanation:

An arch is in the shape of a parabola. It has a span of 100 feet, the vertex lies at the center 50 and the maximum height of 25 ft.

Vertex at (50,25)

vertex form of the equation is

$y=a(x-h)^2+k$ , (h,k) is the center

$y=a(x-50)^2+25$

the parabola starts at (0,0) that is (x,y)

$0=a(0-50)^2+25$

subtract 25 from both sides

$-25=a(-50)^2$

$-25=a(2500)$

divide both sides by 2500

$a=-\frac{1}{100}$

$y=-\frac{1}{100}(x-50)^2+25$

the height of the arch 10 feet from the center.

center is at 50, 10 feet from the center so x=40 and x=60

$y=-\frac{1}{100}(40-50)^2+25$

y=24

the height of the arch 10 feet from the center is 24 feet

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IgorLugansk [536]

14:25
x:y

i hope this helpss

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