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alukav5142 [94]
3 years ago
13

Someone please help me!

Mathematics
1 answer:
Nuetrik [128]3 years ago
5 0

Answer:

33

Step-by-step explanation:

25/30=45/x

x=63

63-30=?

?=33

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HELP ‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎
Simora [160]

Answer:

Aidan

Step-by-step explanation:

Because he interviewed more people.

8 0
3 years ago
Read 2 more answers
Use the substitution u = tan(x) to evaluate the following. int_0^(pi/6) (text(tan) ^2 x text( sec) ^4 x) text( ) dx
Rudiy27
If we use the substitution u = \tan x, then du = \sec^2 {x}\ dx. If you try substituting just u and du into the integrand, though, you'll notice that there's a \sec^2x left over that we have to deal with.

To get rid of this problem, use the identity \tan^2 x + 1 = \sec^2 x and substitute in the left side of the identity for the extra \sec^2x, as shown:

\int\limits^{\pi/6}_0 {tan^2 x \ sec^4 x} \, dx
\int\limits^{\pi/6}_0 {tan^2 x \ (tan^2 x + 1) \ sec^2 x} \, dx

From there, we can substitute in u and du, and then evaluate:

\int\limits^{\pi/6}_0 {tan^2 x \ (tan^2 x + 1) \ sec^2 x} \, dx
\int\limits^{\frac{1}{\sqrt{3}}}_0 {u^2(u^2 + 1)} \, du
\int\limits^{\frac{1}{\sqrt{3}}}_0 {u^4 + u^2} \, du
= \left.\frac{u^5}{5} + \frac{u^3}{3}\right|_0^\frac{1}{\sqrt{3}}
= (\frac{(\frac{1}{\sqrt{3}})^5}{5} + \frac{(\frac{1}{\sqrt{3}})^3}{3}) - (\frac{(0)^5}{5} + \frac{(0)^3}{3})
= \frac{1}{45\sqrt{3}} + \frac{1}{9\sqrt{3}} = \frac{6}{45\sqrt{3}} = \bf \frac{2}{15\sqrt{3}}


8 0
3 years ago
During an experiment the following times (in seconds) were recorded. Find the median.
Debora [2.8K]

Answer:

The median is 2.65 seconds

Step-by-step explanation:

So the median is the middle number. First of all, we have to sort them into ascending order.

2.35, 2.38, 2.43, 2.65, 2.69, 2.78, 10,91.

Since this is median and not mean, we don't need to get rid of outliers / anomalies.

The middle number in that would be 2.65.

6 0
3 years ago
The area of triangle ABC is _____.
AlekseyPX

The formula for the area of a triangle is A = (base · height)/2, but when we look at this problem, there's no base or height!

Looking at the triangle on the right, we can use the Pythagorean Theorem to find the height:

h² + 16² = 20²

h² + 256 = 400

h² = 144

h = ± 12

In this case, the height is 12 because you can't have negatives when you are working with lengths of sides and other things like this in geometry.

So now we've found the height, but we still don't know the full base! So, we can use the height that we just found to use the Pythagorean Theorem for the left triangle!

B² + 12² = 15²

B² + 144 = 225

B² = 81

B = ±9

So once again, our answer is just 9.

But, because the base is CB, we have to add:

b = 9 + 16  

b = 25.

Now we have the base and height, so we can put them into our formula to solve for the area of the triangle:

A = (25 · 12) / 2

A = 300 / 2

A = 150

3 0
3 years ago
Can some one help me with some math problems
Tatiana [17]
It's C, Y = 2x + 3. This is cause, if we use the table and say that y = 5 and x = 1, then 5 = 1 x 2 + 3 which is correct
4 0
3 years ago
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