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3 years ago

5

Because of the commutative property of multiplication, it is true that

1 answer:

dangina [55]3 years ago

8 0

¾ + ¾ + ¾ + ¾ = (3+3+3+3)/4 = 12/4 = 3

¾ × 4 = (3 × 4)/4 = 12/4 = 3

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Solve for n, t=m+n need help to this problem asap thanks ,

loris [4]

T=m+n
m+n=t
n=t-m (In terms of n)

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4 years ago

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Point B has coordinates (3,1). The x-coordinate of point A is −2. The distance between point A and point B is 13 units. What

gogolik [260]

Answer: ask for help on socratic

Step-by-step explanation:

Socratic can help you with any

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3 years ago

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Is 0.48 and 0.52 equally likely or not.

kompoz [17]

Your answer will be D. hope I helped and its the only one that is reasonable

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3 years ago

After refueling, he starts from rest and leaves the pit area with an acceleration whose magnitude is 5.1 m/s2; after 3.6 s he en

Naddik [55]

Answer:

The entering car is going to catch up with the other car after 20.76 seconds.

Step-by-step explanation:

The car leaving the pit area is moving with a constant accelaration, then we can calculate the speed it has when entering the main speedway using the following equation:

$s_1=s_0+a*t_1$

Where $s_0$ is the initial speed (which is zero given that the car starts froms rest), $a$ is the car's accelaration and $t_1$ is the time passed until it reaches the main speedway.

$s_1=s_0+a*t_1=a*t_1$

$s_1=a*t_1=(5.1 \frac{m}{s^2})(3.6s)=18.36 \frac{m}{s}$

For an object travelling at a constant accelation, the displacement $x$ at a time $t$ can be calculate using the following equation:

$x=x_1+(s_1*t)+\frac{1}{2}at^2$ <em>(equation 1)</em>

Let's consider $x_1$ the point where the first car enters the main speedway. If $x_2$ is the point where this car catch up with other one after $t_2$ seconds, we may rewrite the <em>equation 1</em> like this:

$x_2=x_1+(s_1*t_2)+\frac{1}{2}a(t_2)^2$

$x_2=(s_1*t_2)+\frac{1}{2}a(t_2)^2$ <em>(equation 2)</em>

In other hand, the second car is travelling at a constant speed $(v)$ when it meets the entering car. Then, its displacement $(d_2)$ after $t_2$ seconds can be calculated using the following formula:

$d_2=v*t_2$ <em>(equation 3)</em>

The entering car is going to catch up with the other one when $x_2=d_2$ , so when can find how much time this will require equaling <em>equations 2 and 3</em> and isolating $t_2$

$v*t_2=(s_1*t_2)+\frac{1}{2}a(t_2)^2$

$0=(s_1*t_2)+\frac{1}{2}a(t_2)^2-v*t_2$

$[\frac{1}{2}(a*t_2)-(v-s_1)]*t_2=0$

$\left \{ {{t_2=0} \atop {\frac{1}{2}(a*t_2)-(v-s_1)=0}} \right.$

$\frac{1}{2}(a*t_2)-(v-s_1)=0$

$t_2=\frac{2}{a}(v-s_1)=\frac{2}{5.1 \frac{m}{s^2}}(71.3\frac{m}{s}-18.36\frac{m}{s})=20.76 s$

So, the entering car is going to catch up with the other car after 20.76 seconds.

8 0

3 years ago

Which of the following is a solution of z^6 = 64i?

cupoosta [38]

Answer:

2(cos135° + isin135°)

Step-by-step explanation:

z^6 = 64i

We need to change i to polar form

cos (90) + isin (90) = i

x^6 = 64 cos (90) + isin (90)

Now we need to take the sixth root of each side

(x^6) ^ 1/6 = ((64)( cos (90) + isin (90)) ^ 1/6

(x^6) ^ 1/6 = ((64) ^ 1/6 * cos (90) + isin (90)) ^ 1/6

(x^6) ^ 1/6 = 2( * cos (90) + isin (90)) ^ 1/6

We we take the roots of the trig functions, we have 6 roots

360/n means the roots are 60 degrees are apart

take 90 /6 = 15 degrees

The first root is at 2 (cos (15) + isin (15))

The second root is at 2 (cos (15+60) + isin (15+60))

2 (cos (75) + isin (75))

The third root is at 2 (cos (75+60) + isin (75+60))

2 (cos (135) + isin (135))

and so on

6 0

4 years ago