How many seconds are there in (a) one hour and thirty-five minutes and (b) one day?

What even is this i don't even know

mixas84 [53]

The second one.

Hope this helps!!

5 0

3 years ago

System of conics help

ZanzabumX [31]

Answer:

Solutions are (6,3) and (-2,3).

Step-by-step explanation:

From the first equation given we can write:

$(x-2)^{2} = 16-(y-3)^{2}\\$

substituting for $(x-2)^{2}$ in the second equation given we get

$\frac{16-(y-3)^{2}}{16} + \frac{(y-3)^{2}}{64} = 1$

$\frac{-(y-3)^{2}}{16} + \frac{(y-3)^{2}}{64} =0$

∴ y=3

Putting y=3 in the first equation we get

$(x-2)^{2} = 16$

x-2 = ±4

Hence x=6 or x=-2.

7 0

3 years ago

Vijay has a collection of 50 number cubes that are each 1 cubic inch. He finds a box that is 2 inches high. He fits 10 cubes int

galben [10]

Answer:

Vijay cannot fit all the cubes into the box, he can only fit 20 cubes into the box

10 x 2 = 20 cubes

He would need 5 boxes to fit all the cubes

Hope this helps <3

3 0

3 years ago

At 9:00 pm bob leaves his home and starts walking to the bus stop at a constant speed of 9.3 feet per second. bob's house is 432

klio [65]

a. define variables for the quantities that are changing (be specific).
We define variables:
t = time in second
d = distance traveled
The variables that are changing are t and d.
As the time increases, the distance decreases.

b. what does it mean to say bob walks at a constant speed of 9.3 feet per second?
It means that for every second that passes, Bob walks 9.3 feet.

c. consider the formula d = 4320 - 9.3t i. what does t represent? ii. what does 9.3t represent? iii. what does d represent?
For this case we have:
t = represents the time in seconds
9.3t = represents the distance traveled by Bob
d = represents the distance Bob must travel to get to the bus stop.

5 0

3 years ago

Find all solutions of the equation: 2cos^2x-cosx=1

Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

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Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

and the equation becomes

$\mathsf{2t^2-t-1=0}$

Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

3 0

3 years ago