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Korolek [52]

2 years ago

pha \times sin \alpha = " alt=" tan \alpha \times sin \alpha = " align="absmiddle" class="latex-formula">

Mathematics

1 answer:

galina1969 [7]2 years ago

4 0

Tan(α) x sin(α)

Using tan(t) = sin(t) / cos(t), transform the expression:

sin(α) / cos(α) x sin(α)

Calculate the product:

sin(α)² / cos(α)

Final answer:

= sin(α)² / cos(α)

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What is the unit price for 12 apples for $3.12?

horsena [70]

Answer:

$3.85? 12/3.12 = $3.85

The answer would have came out of 3.84615384615

Then round up the answer to 3.85. So the unit price is $3.85.

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2 years ago

Apply the order of operations to simplify each expression.

Nuetrik [128]

Answer:

your right it is 12 great job!

Step-by-step explanation:

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2 years ago

Read 2 more answers

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Anika [276]

Answer:

Step-by-step explanation:

sec (thita) = squrt (5)

squaring on both sides:

sec^2 (thita) = 5 - equation 1

1 + tan^2 (thita) = 5

tan^2 (thita) = 4

tan (thita) = 2.

= tan (thita) - squrt(5)sin(thita)

= 2 - squrt(5) x 2/ squrt(5)

= 0

from eqn - 1

sec(thita) = squrt(5)

cos(thita) = 1/ squrt(5)

sin(thita) = squrt ( 1- 1/(squrt (5))^2)

sin(thita) = 2/ squrt(5) .

4 0

3 years ago

Using mathematical induction prove whether or not the following statement is true for all positive integers n, or show why it is

aniked [119]

Base case: For $n=1$ , the left side is 2 and the right is $2\cdot1^2=2$ , so the base case holds.

Induction hypothesis: Assume the statement is true for $n=k$ , that is

$2+6+10+\cdots+4k-2=2k^2$

We want to show that this implies truth for $n=k+1$ , that

$2+6+10+\cdots+4k-2+4(k+1)-2=2(k+1)^2$

The first $k$ terms on the left reduce according to the assumption above, and we can simplify the $k+1$ -th term a bit:

$\underbrace{2+6+10+\cdots+4k-2}_{2k^2}+4k+2$

$2k^2+4k+2=2(k^2+2k+1)=2(k+1)^2$

so the statement is true for all $n\in\mathbb N$ .

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3 years ago

yan [13]

Answer:

464.5cm³

7 0

2 years ago