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2 years ago

7

Solve t-v/r=k, for v

1 answer:

Arisa [49]2 years ago

6 0

Answer:

t-v/r = k

t-k = v/r

v = r(t-k)

Hope this helps!

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2. Kim is x years old. Jordan is 7 years older than Kim. Four times Jordan’s age is equal to 200. (a) Write an equation that cou

masya89 [10]

Kim is 43 years old so jordan is 50.

200 divided by 4 equals 50.

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3 years ago

Read 2 more answers

Round 27,141 to the nearest ten

Wittaler [7]

27, 140 is the correct answer

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2 years ago

9/11 - 1/2 in it's simplest form

Murrr4er [49]

Answer:

7/22

Step-by-step explanation:

Step 1: <u>Define/explain.</u>

To subtract fractions with different denominators, you must first find the least common denominator (LCM).

You can do this by listing the multiples of each denominator.

Step 2: <u>Solve.</u>

11: 11, 22, 33, 44, 55, 66, 77, 88, 99, 110

2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22

22 is the LCM.

Now, we change the fractions to where they both have denominators of 22.

**REMEMBER: WHICHEVER NUMBER YOU MULTIPLY THE DENOMINATOR BY, YOU MULTIPLY IT BY THE NUMERATOR AS WELL!!**

$9/11*2/2=18/22$

$1/2*11/11=11/22$

Now, subtract as normal.

$18/22-11/22=7/22$

$7/22 (simplified).$

Step 3: <u>Conclude.</u>

Therefore, 9/11 - 1/2 = 7/22.

5 0

3 years ago

I need help I have this problem and I think it either has to be simplified or turned into a fraction but I’m really sure it’s si

Olenka [21]

What are you specifially looking for? simplified it would be 10 and as a fraction 10 over 1

6 0

3 years ago

Find the components of the vertical force Bold Upper FFequals=left angle 0 comma negative 10 right angle0,−10 in the directions

quester [9]

Solution :

Let $v_0$ be the unit vector in the direction parallel to the plane and let $F_1$ be the component of F in the direction of $v_0$ and $F_2$ be the component normal to $v_0$ .

Since, $|v_0| = 1,$

$(v_0)_x=\cos 60^\circ= \frac{1}{2}$

$(v_0)_y=\sin 60^\circ= \frac{\sqrt 3}{2}$

Therefore, $v_0 = \left$

From figure,

$|F_1|= |F| \cos 30^\circ = 10 \times \frac{\sqrt 3}{2} = 5 \sqrt3$

We know that the direction of $F_1$ is opposite of the direction of $v_0$ , so we have

$F_1 = -5\sqrt3 v_0$

$=-5\sqrt3 \left$

$= \left$

The unit vector in the direction normal to the plane, $v_1$ has components :

$(v_1)_x= \cos 30^\circ = \frac{\sqrt3}{2}$

$(v_1)_y= -\sin 30^\circ =- \frac{1}{2}$

Therefore, $v_1=\left< \frac{\sqrt3}{2}, -\frac{1}{2} \right>$

From figure,

$|F_2 | = |F| \sin 30^\circ = 10 \times \frac{1}{2} = 5$

∴ $F_2 = 5v_1 = 5 \left< \frac{\sqrt3}{2}, - \frac{1}{2} \right>$

$=\left$

Therefore,

$F_1+F_2 = \left< -\frac{5\sqrt3}{2}, -\frac{15}{2} \right> + \left< \frac{5 \sqrt3}{2}, -\frac{5}{2} \right>$

$= = F$

3 0

2 years ago