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alexgriva [62]
3 years ago
5

The fourth term in the expansion of the binomial (2x + y2)5 is

Mathematics
1 answer:
Irina-Kira [14]3 years ago
4 0

(2x+y2)5 = 32x5 + 160x4y + 320x3y2 + 320x2y3 + 160xy4 + 32y5

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Please help me!! I need help
Setler [38]

Answer:

c=l/--a^2+b^2=l/--45^2+20^2≈49.24429

Step-by-step explanation:

l/-- mean square root

^ means squared/ exponent

7 0
2 years ago
Read 2 more answers
Can someone please help me find the answers to these two questions please?
Sliva [168]

Answer:

Question 13: x = 47°

Question 14: x = 67°

Step-by-step explanation:

Question 13:

add up all the known factors meaning 16, 27 and 90°, then subtract 180 from what you got from adding known factors

Question 14:

out out of all numbers meaning 19, 4 and 90°, then subtract 180 from what you got from adding all known factors

5 0
3 years ago
Verify the identity. 4 csc 2x = 2 csc2x tan x
vlada-n [284]

Step-by-step explanation:

4csc(2x) = 2csc^2(x) tan(x)

We start with Left hand side

We know that csc(x) = 1/ sin(x)

So csc(2x) is replaced by 1/sin(2x)

4 \frac{1}{sin(2x)}

Also we use identity

sin(2x) = 2 sin(x) cos(x)

4 \frac{1}{2sin(x)cos(x)}

4 divide by 2 is 2

Now we multiply top and bottom by sin(x) because we need tan(x) in our answer

2\frac{1*sin(x)}{sin(x)cos(x)*sin(x)}

2\frac{sin(x)}{sin^2(x)cos(x)}

2\frac{1}{sin^2(x)} \frac{sin(x)}{cos(x)}

We know that sinx/ cosx = tan(x)

Also  1/ sin(x)= csc(x)

so it becomes 2csc^2(x) tan(x) , Right hand side

Hence verified



6 0
3 years ago
7 of 12
astra-53 [7]

Answer:

x = 10

Step-by-step explanation:

Since the parallelograms are similar then the ratios of corresponding sides are equal , that is

=  ( cross- multiply )

6x = 60 ( divide both sides by 6 )

x = 10

Step-by-step explanation:

7 0
2 years ago
Assume that the lines that appear to be tangent are tangent p is the center if each circle find x. ​
Elodia [21]

Answer:

Step-by-step explanation:

Problem One

All quadrilaterals have angles that add up to 360 degrees.

Tangents touch the circle in such a way that the radius and the tangent form a  right angle at the point of contact.

Solution

x + 115 + 90 + 90 = 360

x + 295 = 360

x + 295 - 295 = 360 - 295

x = 65

Problem Two

From the previous problem, you know that where the 6 and 8 meet is a right angle.

Therefore you can use a^2 + b^2 = c^2

a = 6

b =8

c = ?

6^2 + 8^2 = c^2

c^2 = 36 + 64

c^2 = 100

sqrt(c^2) = sqrt(100)

c = 10

x = 10

Problem 3

No guarantees on this one. I'm not sure how the diagram is set up. I take the 4 to be the length from the bottom of the line marked 10 to the intersect point of the tangent with the circle.

That means that the measurement left is 10 - 4 = 6

x and 6 are both tangents from the upper point of the line marked 10.

Therefore x = 6

4 0
3 years ago
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