OH MY GOSH PLSSS HELP!!!

Mrs. Gonzalez bought a puzzle for $10.95, a board game for $15.95, and a table lamp for $17.10. After tax, the total amount due

inn [45]

Answer:

6.5%

Step-by-step explanation:

15.95+10.95+17.10 = 44

46.86-44 = 2.86

2.86/44 = 0.065

0.065 = 6.5%

6 0

3 years ago

The equation f=v+at represents the final velocity of an qbject, f, with an initial velicity, v, and an acceleration rate, a, ove

otez555 [7]

(f-v)/a=t.............

5 0

3 years ago

Read 2 more answers

What is the area of the parallelogram below if b=6.1? Round your answer to the nearest tenth.

Alisiya [41]

The answer to this is 48.8

6 0

3 years ago

the empire state building is 1250 feet tall. IF an object is thrown upward from the top of the building at an initial velocity o

ss7ja [257]

Answer:

Time for the object to get h(max) s = 1.1875 sec

h (max) = 1272.57 feet

Down time for the object to hit the ground = 4.25 sec

Step-by-step explanation:

The relation

h(s) = - 16*s² + 38* s + 1250 (1)

Is equivalent to the equation for vertical shot

Δh = V(i)*t - 1/2g*t² (in this case we don´t have independent term since the shot is from ground level. We can see in (1), the independent term is 1250 feet ( the height of the empire state building), the starting point of the movement.

The description of the movement is:

V(s) = V(i) - g*s ⇒ V(s) = 38 - 32*s

At h(max) V(s) = 0 38/32 = s

So the maximum height is at s = t = 1.1875 sec

The time for the object to pass for starting point is the same

t = 1.1875 sec

h(max) is

h(max) = - 16* (1.1875)² + 38 (1.1875) + 1250

h(max) = - 22,56 + 45.13 + 1250

h(max) = 1272.57 feet

Time for the object to hit the ground is

h(s) = - 1250 feet

-1250 = - 16 s² + 38*s + 1250

-16s² + 38s = 0

s ( -16s + 38 ) = 0

First solution for that second degree equation is x = 0 which we dismiss

then

( -16s + 38 ) = 0 ⇒ 16s = 38 s = 38/16

s = 2.375 sec and we have to add time between h (max) and to get to starting point ( 1. 1875 sec)

total time is = 2.375 + 1.875

Total time = 4.25 sec

4 0

3 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

3 years ago