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2 years ago

15

2 times the quantity 3 times c plus 9

1 answer:

m_a_m_a [10]2 years ago

5 0

2(3c+9).
You are multiplying two by the whole other equation which is the 3c+9 so once you figure that equation out then multiply the whole thing by that two

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The solution is <br><br> u = ?

Tpy6a [65]

Answer:

u = (b/hn)

Step-by-step explanation:

The inverse!!

3 0

1 year ago

What are the vertical asymptote(s) of y= (x-6)/(x+8) (x-7)

White raven [17]

Answer:

x = -8 and x= 7

Step-by-step explanation:

recall that for a rational expression, the vertical asymptotes occur at x-values that causes the expression to become undefined. These occur when the denominator becomes zero.

Hence the asymptototes will occur in x-locations where the denominator , i.e

(x+8)(x-7) = 0

solving this, we get

(x+8) = 0 ----> x = -8

or

(x-7) = 0 ------> x = 7

hence the asymptotes occur x = -8 and x= 7

5 0

2 years ago

Read 2 more answers

Which expression is equivalent to the following complex fraction?

I am Lyosha [343]

what a mess?!

please write your question better...

if you don't want the wrong answer

Step-by-step explanation:

what should we do?

4 0

3 years ago

Translate this phrase into an algebraic expression.

lesya [120]

Answer:

47 - 2n

Step-by-step explanation:

An algebraic expression is an idea of representing numbers using letters or alphabet.

From the question above,

The variable = n

twice the number = 2× n = 2n

Translating:

47 - 2n

8 0

1 year ago

2. When Joey dives off a diving board, the equation of his pathway can be modeled by h = -16- + 15 + 12.

Sergeu [11.5K]

Answer:

a) The maximum height is approx 15.5 unit.

b) The time it will take for Joey to reach the water is 1.45 hour.

Step-by-step explanation:

Given : When Joey dives off a diving board, the equation of his pathway can be modeled by $h(t)= -16t^2+15t + 12$

To find : a) Find Joey's maximum height.

b) Find the time it will take for Joey to reach the water.

Solution :

Modeled $h(t)= -16t^2+15t + 12$ ....(1)

a) To find maximum height

Derivate (1) w.r.t. t,

$h'= -32t+15$

For critical point put h'=0,

$-32t+15=0$

$t=\frac{15}{32}$

$t=0.46875$

Derivate again w.r.t. t,

$h''= -32$

It is maximum at t=0.46875.

Substitute t in equation (1),

$h(0.46875)= -16(0.46875)^2+15(0.46875)+12$

$h(0.46875)= -3.515625+7.03125+12$

$h(0.46875)= 15.515625$

The maximum height is approx 15.5 unit.

b) To find the time it will take for Joey to reach the water.

Put h=0 in equation (1),

$-16t^2+15t + 12=0$

Apply quadratic formula, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Here, a=-16 , b=15, c=12

$t=\frac{-15\pm\sqrt{(15)^2-4(-16)(12)}}{2(-16)}$

$t=\frac{-15\pm\sqrt{225+768}}{-32}$

$t=\frac{-15\pm\sqrt{993}}{-32}$

$t=\frac{-15+\sqrt{993}}{-32},\frac{-15-\sqrt{993}}{-32}$

$t=−0.515,1.453$

Reject negative value.

The time is t=1.45.

The time it will take for Joey to reach the water is 1.45 hour.

5 0

3 years ago