15. Find the values of x and y.<br> (10x - 61)°<br> (18y + 5)º<br> (x + 10)

Choose the correct Set-builder form for the following set written in Roster form: { − 8 , − 7 , − 6 , − 5 , − 4 }

PIT_PIT [12445]

Response:

Detailed explanation:

5 0

3 months ago

Let c1(t) = eti + (sin(t))j + t3k and c2(t) = e−ti + (cos(t))j − 6t3k. Find the stated derivatives in two different ways to veri

Zina [12379]

Answer:

$i ( e^{t} - e^{-t})+ j (cost-sin t)+ k (-15t^{2})$

$\frac{d}{dx}(e^x) = e^x$

Step-by-step explanation:

Step 1:-

We have c1(t) = e^ t i + (sin(t))j + t³k

and c2(t) = e^−t i + (cos(t))j − 6t³k.

By adding c1(t) and c2(t):

c1(t)+c2(t) = e^ t i + (sin(t))j + t³k + e^−t i + (cos(t))j − 6t³k

Now, employing the derivative formula:

$\frac{d}{dx}(e^x) = e^x$

$\frac{d}{dx}(sinx) = cosx\\\frac{d}{dx}(cosx) = -sinx$

Next, differentiate with respect to 't'

$\frac{d}{dt}c_{1}+ c_{2} } = e^ t i +cost j +3t^2 k - e^-t i - sintj -18t^2 k$

By factoring out i, j, and k terms, we arrive at:

$\frac{d}{dt}(C_{1} +C_{2} ) = i ( e^{t} - e^{-t})+ j (cost-sin t)+ k (-15t^{2})$

7 0

3 months ago

3. A CD costs £9.50 in London and

PIT_PIT [12445]

Detailed explanation:

Question 4 provides clear information,

thus resolving Question 4:

1 euro equals $1.17.

Consequently, €20.46 is calculated as 20.46 multiplied by 1.17, which equals $23.93.

1 pound equals $1.28.

Thus, £12.60 translates to 12.60 multiplied by 1.28, yielding $16.128.

Therefore, the MP3 in pounds is $7.81 cheaper in dollars.

6 0

3 months ago

A baseball is thrown up in the air. The table shows the heights y (in feet) of the baseball after x seconds.

lawyer [12517]

Answer:

Alright, we can express the baseball's motion with an equation like:

h(x) = a*x^2 + b*x + c

Here, x denotes time, while h(x) indicates height.

Let’s construct this:

The acceleration is:

a(t) = a

For velocity, integrating over time results in:

v(x) = a*x + v0

Where v0 signifies the initial vertical velocity.

Subsequently, we can determine position or height by integrating once more:

h(x) = a*x^2 + v0*x + h0

Here, h0 is the initial height.

h(x) = a*x^2 + v0*x + h0.

When x = 0s, h(0s) = 6ft

<pthus:>

h(0s) = a*0s^2 + v0*0s + h0 = 6ft

h0 = 6ft.

It’s also noted that:

h(2s) = h(4s)

a*x^2 + b*x + c

The symmetry line is given by:

x = -b/2a

b = v0

a = a

3s = -v0/(2*a)

v0 = -3s*(2a)

h(x) = a*x^2 - 6s*a*x + 6ft

h(2s) = 22ft = a*(2s)^2 - 6s*a*2s + 6ft

h(x) = (-2ft/s^2)*x^2 + (12ft/s)*x + 6ft

b) The height after 5 seconds is expressed as:

h(5s) = (-2ft/s^2)*(5s)^2 + (12ft/s)*5s + 6ft = 16ft

</pthus></p></pso:></pnext></phaving></ptherefore></pin></pin></pthe></pthus:></pexamining></pthus>

7 0

4 months ago

15. Find the values of x and y. (10x - 61)° (18y + 5)º (x + 10)