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1 year ago

Can you help me18f=-216

1 answer:

3 0

$\begin{gathered} 18f=-216 \\ \text{Solve for f:} \\ \text{Divide both sides by 18:} \\ \frac{18f}{18}=-\frac{216}{18} \\ f=-12 \end{gathered}$

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Find an equation of combined variation where a varies directly as b and inversely as c. One set of values is a = 4, b = 12, and

Ber [7]

Answer:

Combined equation is $a = \frac{kb}{c}$

Step-by-step explanation:

a varies directly as b and inversely as c.

This can be written as

a = $k \times b$

a = kb

where k is the proportionality constant

$a = \frac{kb}{c}$ -------------------------------------(1)

Now lets find the k value bu substituting the given a, b,c values

$4 = \frac{k (12)}{9}$

$9 \times 4 = 12k$

36 = 12 k

$k = \frac{36}{12}$

k = 3

Thus the eq(1) becomes

$a = \frac{3b}{c}$

Let us now find the value of a when b=7 and c = 3

$a = \frac{3(7)}{3}$

$a =\frac{21}{7}$

a = 7

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4 years ago

What's negative five twelfths minus three tenths?

Scorpion4ik [409]

-0.7166666666666666 that is the answer I got but can be rounded to -0.72

3 0

3 years ago

Read 2 more answers

A baseball pitcher won 60% of his games. How many did he win if he pitched 35 games?

Sergio039 [100]

Answer:

21 games

Step-by-step explanation:

Multiply the number of games by the percentage.

35 · 0.60 = 21

6 0

3 years ago

What is measure of side A?

Ira Lisetskai [31]

<u>Answer:</u>

The correct answer option is c. a = 11.94.

<u>Step-by-step explanation:</u>

We are given a right angles triangle with an angle ABC equal to 58 degrees and side length AC known to be equal to 19.1.

We are supposed to find the length of side a.

Side AC, opposite to angle ABC, will be the perpendicular and BC will be the base so we will use the formula of tan to find b.

$tan 58 = \frac{19.1}{a}$

$a = \frac{19.1}{tan 58}$

$a=11.94$

8 0

4 years ago

A child wanders slowly down a circular staircase from the top of a tower. With x,y,zx,y,z in feet and the origin at the base of

babymother [125]

Answer:

a) The tower is 90 feet tall

b) She reaches the bottom at t = 18 minutes.

c) Her speed at time t is 5 \sqrt[]{5} ft/minute

d) Her acceleration at time t is 10 ft/minute^2

Step-by-step explanation:

Consider the path described by the child as going down the tower to have the following parametrization $\gamma(t) = (10\cos t, 10 \sin t, 90-5t)$

a) Assuming that the child is at the top of the tower when she starts going down, we have that at the initial time (t=0) we will have the value of the height of the tower. That is z = 90-5*0 = 90 ft.

b) The child reaches the bottom as soon as z =0. We want to find the value of t that does that. Then we have 0 = 90-5t, which gives us t = 18 minutes.

c) Given the parametrization we are given, the velocity of the child at time t is given by $\frac{d\gamma}{dt}= (\frac{d}{dt}(10\cos t), \frac{d}{dt} (10 \sin t ), \frac{d}{dt}(90-5t)) = (-10 \sin t, 10 \cos t, -5)$ . The speed is defined as the norm of the velocity vector,

so, the speed at time t is given by $v = \sqrt[]{(-10 \sin t)^2+(10 \cos t)^2+(-5)^2} = \sqrt[]{100(\sin^2 t + \cos^2 t)+25} = \sqrt[]{125}= 5 \sqrt[]{5}$

d) ON the same fashion we want to know the norm of the second derivative of $\gamma$ .

We have that $\gamma ^{''}(t) =(-10\cost t, -10 \sin t , 0)$ so the acceleration is given by $\sqrt[]{100(\cos^2 t+ \sin^2 t )} = 10$

6 0

3 years ago