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Dominik [7]
3 years ago
15

1/(cot^2 (x)) - 1/(cos^2 (x))

Mathematics
2 answers:
Airida [17]3 years ago
5 0

Step-by-step explanation:

1 is your answer its law of trig identises

tia_tia [17]3 years ago
5 0

\\ \sf\longmapsto \dfrac{1}{cot^2x}-\dfrac{1}{cos^2x}

\\ \sf\longmapsto tan^2x-sec^2x

\\ \sf\longmapsto 1

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Which is the slope of a line that is parallel to the line represented by the equation x-y=8
pychu [463]

Answer:

slope of parral line is same as slope of line its 1

5 0
3 years ago
Read 2 more answers
PLEASEEEEE HELPPPP MEEEEEEE BRAINLIEST TO THE FIRST ANSWER AND 30 POINTS
icang [17]

Answer:

CD ≠ EF

Step-by-step explanation:

Using the distance formula

d = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2    }

with (x₁, y₁ ) = C(- 2, 5) and (x₂, y₂ ) = D(- 1, 1)

CD = \sqrt{(-1+2)^2+(1-5)^2}

      = \sqrt{1^2+(-4)^2}

      = \sqrt{1+16} = \sqrt{17}

Repeat using (x₁, y₁ ) = E(- 4, - 3) and (x₂, y₂ ) = F(- 1, - 1)

EF = \sqrt{(- 1+4)^2+(-1+3)^2}

     = \sqrt{3^2+2^2}

     = \sqrt{9+4} = \sqrt{13}

Since \sqrt{17} ≈ \sqrt{13} , then CD and EF are not congruent

     

6 0
3 years ago
What is the value of x?<br> 38<br> 21<br> 21<br> Drawing not to scale
Snezhnost [94]
Where is it?
step by step:
3 0
3 years ago
PLEASE HELP ASAP WILL MARK THE BRAINLEST
statuscvo [17]

Answer:

similar triangles

Step-by-step explanation:

First of all, what are similar shapes? Well, two shapes are similar if you can turn one into the other by moving, rotating, flipping, or scaling. That means you can make one shape bigger or smaller. In this case, we know that triangles ABC and DEF are mathematically similar. The area of triangles ABC is , so we need to know the area of triangle DEF.

From math, let's call  the scaling factor, so we know that for any similar figures, the ratio of the areas of any are in proportion to . In other words, if  is the area of triangle ABC, and  is the area of triangle DEF, then we can write the following relationship:

4 0
3 years ago
Type the correct answer in each box. If necessary, use / for the fraction bar(s).
Tom [10]

Given:

The values are a=0.\overline{6},b=0.75.

To find:

The values of ab and \dfrac{a}{b}.

Solution:

We have,

a=0.\overline{6}

a=0.666...                     ...(i)

Multiply both sides by 10.

10a=6.666...                 ...(ii)

Subtracting (i) from (ii), we get

10a-a=6.666...-0.666...

9a=6

a=\dfrac{6}{9}

a=\dfrac{2}{3}

And,

b=0.75

b=\dfrac{75}{100}

b=\dfrac{3}{4}

Now, the product of a and b is:

ab=\dfrac{2}{3}\times \dfrac{3}{4}

ab=\dfrac{1}{2}

The quotient of a and b is:

\dfrac{a}{b}=\dfrac{\dfrac{2}{3}}{\dfrac{3}{4}}

\dfrac{a}{b}=\dfrac{2}{3}\times \dfrac{4}{3}

\dfrac{a}{b}=\dfrac{8}{9}

Therefore, the required values are ab=\dfrac{1}{2} and \dfrac{a}{b}=\dfrac{8}{9}.

6 0
3 years ago
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