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3 years ago

A rain gutter is made from aluminum sheets that

Mathematics

1 answer:

netineya [11]3 years ago

4 0

Answer:

Step-by-step explanation:

For any quadratic function f(x)=ax²+bx+c given in form f(x)=a(x-h)²+k, where:

h=-b/2a and k=f(h) the point (h, k) is vertex of the function graph.

It means that k is maximum (if a<0) or minimum (if a>0) value of function and x=h is the x which will maximize function.

A(x) = x(28-2x) = 28x - 2x² = - 2x² + 28x ⇒ a = -2 , b = 28

h = -28/[2•(-2)] = 7

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Lynna [10]

Answer:

The formula is 10 - 3n

Step-by-step explanation:

For an nth term in an arithmetic sequence

U( n ) = a + ( n - 1)d

Where n is the number of terms

a is the first term

d is the common difference

From the sequence above

a = 7

d = 4 - 7 = - 3

The formula for an nth term is

U(n) = 7 + (n - 1)-3

= 7 - 3n + 3

The final answer is

= 10 - 3n

Hope this helps you.

3 0

2 years ago

What is the main intent of this report? to explain the services of Fast Pax to convince people to ship using Fast Pax to convinc

AleksAgata [21]

The third option is the answer

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Condense the expression to the logarithm of a single quantity.<br><br>

vladimir1956 [14]

Answer:

$\log _3\left(\frac{x^{\frac{1}{2}}}{\left(y+8\right)^2}\right)$

Step-by-step explanation:

-> Apply log rules

$\log _3\left(x^{\frac{1}{2}}\right)-2\log _3\left(y+8\right)$

$\log _3\left(x^{\frac{1}{2}}\right)-\log _3\left(\left(y+8\right)^2\right)$

-> Apply basic math rules

$\log _3\left(\frac{x^{\frac{1}{2}}}{\left(y+8\right)^2}\right)$

7 0

1 year ago

Simplify the expression: 5^2 - 3 * 6 A)7 B)24 C)132 D)144

kow [346]

Answer:

Step-by-step explanation:

5^2 = 5 x 5 = 25

3 x 6 is 18

25 - 18 is 7

also, use PEMDAS

5 0

3 years ago

Read 2 more answers

If 2^n + 1 is an odd prime for some integer n, prove that n is a power of 2. (H

vovikov84 [41]

Step-by-step explanation:

We will prove by contradiction. Assume that $2^n + 1$ is an odd prime but n is not a power of 2. Then, there exists an odd prime number p such that $p\mid n$ . Then, for some integer $k\geq 1$ ,

$n=p\times k.$

Therefore

$2^n + 1=2^{p\times k} + 1=(2^{k})^p + 1^p.$

Here we will use the formula for the sum of odd powers, which states that, for $a,b\in \mathbb{R}$ and an odd positive number $n$ ,

$a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2-...+b^{n-1})$

Applying this formula in 1) we obtain that

$2^n + 1=2^{p\times k} + 1=(2^{k})^p + 1^p=(2^k+1)(2^{k(p-1)}-2^{k(p-2)}+...-2^{k}+1)$ .

Then, as $2^k+1>1$ we have that $2^n+1$ is not a prime number, which is a contradiction.

In conclusion, if $2^n+1$ is an odd prime, then n must be a power of 2.

4 0

3 years ago