A1 = 5 and d = 5 find the 25th term

Mathematics

1 answer:

boyakko [2]3 years ago

8 0

An = a + (n - 1)d
A25 = -1 + (24)-10
A25 = -1 -240
A25 = -241

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bazaltina [42]

The answer is 21.17 or 21.16601049.

8 0

3 years ago

The diameter of a large sphere is 4 times the diameter of a small sphere. The surface area of the large sphere is how many times

Sever21 [200]

Surface Area of a Sphere = 4π(d/2)²
Let the smaller sphere's diameter be just a sample of 4 units in diameter and let the large spheres diameter be 4 times that of the smaller sphere.

Small Sphere: 4π(4/2)² = 16π units²
Large Sphere: 4π(16/2)² = 256π units²

256π / 16π = 16 times

The large sphere has a surface area 16 times that of the smaller one.

4 0

3 years ago

What is the coefficient of x2y3 in the expansion of (2x + y)5?

Zigmanuir [339]

Option C:

The coefficient of $x^{2} y^{3}$ is 40.

Solution:

Given expression:

$(2 x+y)^{5}$

Using binomial theorem:

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here $a=2 x, b=y$

Substitute in the binomial formula, we get

$(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}$

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}$