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3 years ago

HEY CAN ANYONE PLS ANSWER DIS MATH QUESTION!!!

Mathematics

1 answer:

mrs_skeptik [129]3 years ago

5 0

Answer:

See below.

Step-by-step explanation:

Just make a line that doesn't have a pattern like 2, 4, 6, 8 or like 6, 12, 18, 24.

99.9% sure just spamming the keyboard with random numbers will work.

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5 In a scale drawing of an airplane, 1 inch equals 3 meters. The length of the actual plane is 17 meters. What is the length of

Whitepunk [10]

Answer: 5 2/3 inches

Step-by-step explanation:

From the question, we are informed that the scale drawing of an airplane is:

1 inch = 3 meters.

Since the length of the actual plane is 17 meters, the length of the plane in the scale drawing will then be:

= 17/3

= 5 2/3 inches

4 0

3 years ago

Your credit card has a balance of $5400 and an annual interest rate of 17%. You decide to pay off the balance over two years. If

Phoenix [80]

Answer:

5400/12= 450 paid

1006.56/2=503.28 interest

total is 5,903.28

total for paid 491.94 for months

Step-by-step explanation:

6 0

4 years ago

A 10-page play is $0.70 per page to e

Bas_tet [7]

Answer:

The answer of this question is 7

Step-by-step explanation:

because 1 page=$0.70 and now calculate

10×1×0.70=7

8 0

2 years ago

Module 07: Project Option 1

Flauer [41]

Where are the questions

4 0

3 years ago

A particle moves with velocity vector

asambeis [7]

Answer:

$\vec{s(t)}=\frac{1}{2}t^2\hat{i}-(2t+1)\hat{j}+(\frac{1}{3}t^3+1)\hat{k}$

Step-by-step explanation:

We are given that velocity vector of a particle

$\vec{v(t)}=t\hat{i}-2\hat{j}+t^2\hat{k}$

When t=0 then the particle is at the point (0,-1,1).

We have to find the position of particle at time t.

We know that

Velocity = $\frac{Displacement }{time}=\frac{ds}{dt}$

Therefore, $\vec{v}=\frac{\vec{ds}}{dt}$

$\int{ds}=\int (t\hat{i}-2\hat{j}+t^2\hat{k})dt$

Integrate on both sides then we get

$\vec{s(t)}=\frac{1}{2}t^2\hat{i}-2t\hat{j}+\frac{1}{3}t^3\hat{k}+C$

$\int x^n dx=\frac{x^{n+1}}{n+1}+C$

Substitute the value of point at time t=0 then we get

C= $-\hat{j}+\hat{k}$

Substitute the value of C then we get

$\vec{s(t)}=\frac{1}{2}t^2\hat{i}-2t\hat{j}+\frac{1}{3}t^3\hat{k}-\hat{j}+\hat{k}$

Therefore, the position of particle at time t

$\vec{s(t)}=\frac{1}{2}t^2\hat{i}-(2t+1)\hat{j}+(\frac{1}{3}t^3+1)\hat{k}$

8 0

3 years ago