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3 years ago

Evaluate 3x+5(y+3) when x is 2 less than the quotient of y and 3, and y=12

Mathematics

1 answer:

myrzilka [38]3 years ago

8 0

Answer:

Step-by-step explanation:

12/3 = 4

4 - 2 = 2

x = 2

using the order of operations, you would first have to do y + 3

y = 12

12 + 3 = 15

the equation simplified is 3 x 2 + 5 x 15

3 x 2 = 6

5 x 15 = 75

75 + 6 = 81

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What is sin 45°?<br> Lol I need help

Dahasolnce [82]

Answer:

1/ √2

Step-by-step explanation:

Here, we want to get the value of sin 45

Mathematically the sine is the ratio of the opposite to the hypotenuse

The opposite here is 1 while the hypotenuse is

√2

so the value of sin 45 will be;

1/ √2

6 0

3 years ago

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

Isaiah walks 3 miles due north, turns and then 5 miles due east. How far is he from his starting point?

Sonbull [250]

Isiah is 5.83 miles away from his starting point.

When you draw the the miles Isaiah walks, you will discover that you are forming a right angle triangle.

To measure the length from the starting point to the ending point. We must used the Pythagorean equation. Where in the square of the hypotenuse is equal to the sum of squares of the other two sides.

a² + b² = c²

a = 3 miles 3² + 5² = c²
b = 5 miles 9 + 25 = c²
c = hypotenuse. 34 = c²

We must get the hypotenuse, since 34 is the square of the hypotenuse we must look for its square root.

c² = 34
c = √34
c = 5.83 miles

7 0

3 years ago

What is the following sum?<br> 4(^5sqrtx^2y) + 3(^5sqrtx^2y)

Schach [20]

Answer:

option 3

Step-by-step explanation:

$(4 + 3) \sqrt[5]{x {}^{2}y } = 7 \sqrt[5]{ {x}^{2}y }$