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grigory [225]
3 years ago
14

I need help finding v

Mathematics
1 answer:
Anvisha [2.4K]3 years ago
5 0
Check the picture below.

surely you can solve for "v", right?

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Put a digit in the gap, such that the answer is a whole number<br><br> 10x+11=144...
d1i1m1o1n [39]

Answer:

the answer is 1

Step-by-step explanation:

If you do 1441-11 you get 1430.

no other number gives you 0.

8 0
3 years ago
One of the legs of a right triangle measures 12 cm and the other leg measures 13 cm find the measure of the hypotenuse if necess
Amanda [17]

Answer:

17.7 cm

Step-by-step explanation:

One of the legs of a right triangle measures 12 cm and the other leg measures 13 cm find the measure of the hypotenuse if necessary round the nearest 10th

To find the Hypotenuse of a right angle triangle, we solve using Pythagoras Theorem

Hypotenuse ² = Opposite ² + Adjacent ²

Hypotenuse = √Opposite ² + Adjacent ²

Opposite = 12 cm

Adjacent = 13 cm

Hence,

Hypotenuse = √12² + 13²

= √144 + 169

= 17.691806013 cm

Approximately = 17.7 cm

Therefore, the measure of the Hypotenuse is 17.7 cm

6 0
3 years ago
What is the solution to the equation 5x = 40?
Dima020 [189]
To find X, we can divide 5 into 40.
40 divided by 5 is 8.
Now we can do 5 x 8. That is 40.

X=40

OR we can do it another way.
40 divided by 5 is 8.
Cross out 5 divided by 5.

Now we have 8. Still.
X is 8
7 0
2 years ago
Read 2 more answers
Find derivative problem<br> Find B’(6)
dalvyx [7]

Answer:

B^\prime(6) \approx -28.17

Step-by-step explanation:

We have:

\displaystyle B(t)=24.6\sin(\frac{\pi t}{10})(8-t)

And we want to find B’(6).

So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:

\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)]

We can move the constant outside:

\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]

Now, we will utilize the product rule. The product rule is:

(uv)^\prime=u^\prime v+u v^\prime

We will let:

\displaystyle u=\sin(\frac{\pi t}{10})\text{ and } \\ \\ v=8-t

Then:

\displaystyle u^\prime=\frac{\pi}{10}\cos(\frac{\pi t}{10})\text{ and } \\ \\ v^\prime= -1

(The derivative of u was determined using the chain rule.)

Then it follows that:

\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}

Therefore:

\displaystyle B^\prime(6) =24.6[(\frac{\pi}{10}\cos(\frac{\pi (6)}{10}))(8-(6))- \sin(\frac{\pi (6)}{10})]

By simplification:

\displaystyle B^\prime(6)=24.6 [\frac{\pi}{10}\cos(\frac{3\pi}{5})(2)-\sin(\frac{3\pi}{5})] \approx -28.17

So, the slope of the tangent line to the point (6, B(6)) is -28.17.

5 0
3 years ago
Simplify: −z^3+5k^6−(−z^3+10k^6)
Georgia [21]
-z³ + 5k^6 + z³ -10k^6

(-z³ cancels out with z³)

5k^6 -10k^6

(then subtract)

Answer is -5k^6
6 0
3 years ago
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