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Norma-Jean [14]

3 years ago

10

Write an equation of the line passing through point p(−1, 3) that is perpendicular to the line x = −5.

1 answer:

Crank3 years ago

4 0

The answer is y=1/5x-14/5

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Use the rules for long division to divide 262 by 9.

TEA [102]

______2___9_______
9| 2 6<span> </span>2
-1 8
---------
8 2
-1 8
----------
1

So, 29 with a remainder of 1, meaning the answer is A.

3 0

4 years ago

Read 2 more answers

A 15 kilogram object is suspended from the end of a vertically hanging spring stretches the spring 1/3 meters. At time t = 0, th

Yuri [45]

Answer:

$15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) =$ ± $170 cos(5t)$

$y(0)=0$ , $y'(0)=0$

Step-by-step explanation:

See the attached image

This problem involves Newton's 2nd Law which is: ∑F = ma, we have that the acting forces on the mass-spring system are: $F_{r} (t)$ that correspond to the force of resistance on the mass by the action of the spring and $F(t)$ that is an external force with unknown direction (that does not specify in the enounce).

For determinate $F_{r} (t)$ we can use Hooke's Law given by the formula $F_{r} (t) = k y(t)$ where $k$ correspond to the elastic constant of the spring and $y(t)$ correspond to the relative displacement of the mass-spring system with respect of his rest state.

We know from the problem that an 15 Kg mass stretches the spring 1/3 m so we apply Hooke's law and obtain that...

$k = \frac{F_{r}}{y} = \frac{mg}{y} = \frac{15 Kg (9.81 \frac{m}{s^{2} } )}{\frac{1}{3} m} = 441.45 \frac{N}{m}$

Now we apply Newton's 2nd Law and obtaint that...

$F_{r} (t)$ ± $F(t)$ = $ma(t)$

$F_{r} (t) = ky(t) = 441.45y(t)$

$F(t) = 170 cos(5t)$

$m = 15 kg$

$a(t) = \frac{d^{2}y(t)}{dt^{2} }$

Finally... $15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) =$ ± $170 cos(5t)$

We know from the problem that there's not initial displacement and initial velocity, so... $y(0)=0$ and $y'(0)=0$

Finally the Initial Value Problem that models the situation describe by the problem is

$\left \{ 15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) = \frac{+}{} 170 cos(5t) \atop {y(0)=0, y'(0)=0\right.$

6 0

3 years ago

Select Transform to change the drawing of the figure. What is the area of the front of the prism? Area = cm2 2 cm 3 cm 6 cm Tran

sasho [114]

Answer:12 cm2

Step-by-step explanation:I’m smart

4 0

3 years ago

Read 2 more answers

Let y1, y2,... , yn denote a random sample of size n from a population with a uniform distribution on the interval (0,theta). le

laila [671]

Step-by-step explanation:

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6 0

3 years ago

A pyramid has a square base and a height of 6 ft. The volume of the pyramid is 162 ft^3. Let s be the length of a side of the py

geniusboy [140]

Answer:

$(a)s^2\\(b)162=2s^2\\(c)s=9$ ft$

Step-by-step explanation:

(a)Base Area

Side length of square base=s

Area of a Square of side length s= $s^2$ ft^2$

(b)

Height =6 ft
Volume= $162 $ ft^3.$

Volume of a Pyramid $=\frac{1}{3}X$Base Area X Height$

Substituting the given values, we have:

$162=\frac{1}{3}Xs^2 X 6\\\\162=2s^2$

(c)Solving the equation derived from (b)

$162=2s^2\\$Divide both sides by 2\\s^2=81\\s^2=9^2\\s=9$ ft$

Solving this equation shows that s =9 ft.

4 0

3 years ago