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3 years ago

PROBLEM 2 What is the greatest common factor of 15 and 30?

Mathematics

2 answers:

seropon [69]3 years ago

7 0

Answer: 15

Step-by-step explanation:

Liono4ka [1.6K]3 years ago

4 0

Answer:

$15 = 3 \times 5 \\ 30 = 3 \times 5 \times 2 \\ \\ \: \: \: \: \: \: \: = 3 \times 5 \\ \: \: \: \: \: \: \: = 15$

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Find the area of the shape shown below.

Igoryamba

Answer:

Area = 16

Step-by-step explanation:

4 x 2 x 2 x 1 = 16

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3 years ago

Please help i’m struggling so much

Lesechka [4]

3 should be the answer

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3 years ago

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Complete the identity. 1) sec^4 x + sec^2 x tan^2 x - 2 tan^4 x = ?

Alecsey [184]

Answer:

See Explanation

Step-by-step explanation:

Question like this are better answered if there are list of options; However, I'll simplify as far as the expression can be simplified

Given

$sec^4 x + sec^2 x tan^2 x - 2 tan^4 x$

Required

Simplify

$(sec^2 x)^2 + sec^2 x tan^2 x - 2 (tan^2 x)^2$

Represent $sec^2x$ with a

Represent $tan^2x$ with b

The expression becomes

$a^2 + ab- 2 b^2$

Factorize

$a^2 + 2ab -ab- 2 b^2$

$a(a + 2b) -b(a+ 2 b)$

$(a -b) (a+ 2 b)$

Recall that

$a = sec^2x$

$b = tan^2x$

The expression $(a -b) (a+ 2 b)$ becomes

$(sec^2x -tan^2x) (sec^2x+ 2 tan^2x)$

..............................................................................................................................

In trigonometry

$sec^2x =1 +tan^2x$

Subtract $tan^2x$ from both sides

$sec^2x - tan^2x =1 +tan^2x - tan^2x$

$sec^2x - tan^2x =1$

..............................................................................................................................

Substitute 1 for $sec^2x - tan^2x$ in $(sec^2x -tan^2x) (sec^2x+ 2 tan^2x)$

$(1) (sec^2x+ 2 tan^2x)$

Open Bracket

$sec^2x+ 2 tan^2x$ ------------------This is an equivalence

$(secx)^2+ 2 (tanx)^2$

Solving further;

................................................................................................................................

In trigonometry

$secx = \frac{1}{cosx}$

$tanx = \frac{sinx}{cosx}$

Substitute the expressions for secx and tanx

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$(secx)^2+ 2 (tanx)^2$ becomes

$(\frac{1}{cosx})^2+ 2 (\frac{sinx}{cosx})^2$

Open bracket

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

$\frac{1}{cos^2x}+ \frac{2sin^2x}{cos^2x}$

Add Fraction

$\frac{1 + 2sin^2x}{cos^2x}$ ------------------------ This is another equivalence

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In trigonometry

$sin^2x + cos^2x= 1$

Make $sin^2x$ the subject of formula

$sin^2x= 1 - cos^2x$

................................................................................................................................

Substitute the expressions for $1 - cos^2x$ for $sin^2x$

$\frac{1 + 2(1 - cos^2x)}{cos^2x}$

Open bracket

$\frac{1 + 2 - 2cos^2x}{cos^2x}$

$\frac{3 - 2cos^2x}{cos^2x}$ ---------------------- This is another equivalence

8 0

3 years ago

How many degrees is 4 9 radians?

melomori [17]

80 hope you have a great day

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3 years ago

Help please please $)

mart [117]

<h3>Look at the picture.</h3>

commutative property a + b = b + a

... (-5 + 1/2 + 5) = ... (-5 + 5 + 1/2)

inverse property of addition a + (-a) = a - a = 0

... (-5 + 5 + 1/2) = ... (0 + 1/2)

identity property of addition a + 0 = 0 + a = a

... (0 + 1/2) = ... 1/2

inverse property of multiplication a/b · b/a = 1

37 · 2 · 1/2 = 37 · 2/1 · 1/2 = 37 · 1

identity property of multiplication a · 1 = 1 · a = a

37 · 1 = 37

6 0

3 years ago