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andreyandreev [35.5K]
3 years ago
5

Find the value of x round to the nearest tenth

Mathematics
1 answer:
Leto [7]3 years ago
4 0

Answer:

58.6

Step-by-step explanation:

Use the formula c=\sqrt{a^{2}+b^{2}

solving for x (b)

b=\sqrt{c^{2}-a^{2} = \sqrt{61^{2}-17^{2} ≈58.58327

and 58.58327 rounds to 58.6

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Four less than the square of a number is 0​
marusya05 [52]

Answer:

± 2

Step-by-step explanation:

let the number be n , then

n² - 4 = 0 ( add 4 to both sides )

n² = 4 ( take square root of both sides )

n = ± \sqrt{4} = ± 2

The number is - 2 or 2

5 0
3 years ago
Read 2 more answers
The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in s
Norma-Jean [14]

The positions when the particle reverses direction are:

s(t_1)=55ft\\\\s(t_2)=28ft

The acceleraton of the paticle when reverses direction is:

a(t_1)=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=18\frac{ft}{s^{2}}

Why?

To solve the problem, we need to remember that if we derivate the position function, we will get the velocity function, and if we derivate the velocity function, we will get the acceleration function. So, we will need to derivate two times.

Also, when the particle reverses its direction, the velocity is equal to 0.

We are given the following function:

s(t)=2t^{3}-21t^{2}+60t+3

So,

- Derivating to get the velocity function, we have:

v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60

Now, making the function equal to 0, to find the times when the particle reversed its direction, we have:

v(t)=6t^{2}-42t+60\\\\0=6t^{2}-42t+60\\\\0=t^{2}-7t+10\\(t-5)*(t-2)=0\\\\t_{1}=5s\\t_{2}=2s

We know that the particle reversed its direction two times.

- Derivating the velocity function to find the acceleration function, we have:

a(t)=\frac{dv}{dt}=6t^{2}-42t+60\\\\a(t)=12t-42

Now, substituting the times to calculate the accelerations, we have:

a(t_1)=a(2s)=12*2-42=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=12*5-42=18\frac{ft}{s^{2}}

Now, substitutitng the times to calculate the positions, we have:

s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft

Have a nice day!

3 0
3 years ago
Is this equation linear: 2x+5xy-3=0
Lina20 [59]

What is linear equation?

An equation between two variables that gives a straight line when plotted on a graph.

2x + 5xy - 3 = 0

x (2 + 5y) -3 = 0

x ( 2 + 5y ) = 3

2 + 5y = 3 / x

5y = (3/x) - 2

y = ( (3/x) - 2) / 5

According to the graph i don't think it's a linear equation.

3 0
3 years ago
What is the solution to the equation below?<br> 12 + square root 1 - 5x = 18
GrogVix [38]

Answer:

-1

Step-by-step explanation:

Square root 1 is 1. Add that to the 12 to get 13. Subtract 13 from both sides to get: -5x = 5. Divide both sides by -5 to get your answer: -1

5 0
3 years ago
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Find the slope and y-intercept of the line 6x−7y=2
Allisa [31]
To easily find the slope and y intercept of the line, put it in slope intercept form by rearranging the variables around.

6x-7y = 2
6x - 7y + 7y = 2 + 7y
6x = 7y + 2
6x - 2 = 7y + 2 - 2
7y = 6x - 2
7y/7 = 6x/7 - 2/7
Y = 6/7(X) - 2/7

Y = 6/7x - 2/7
Slope = m = 6/7
Y intercept = b = -2/7
(0, -2/7).
3 0
3 years ago
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