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2 years ago

Need help with this

Mathematics

2 answers:

Marta_Voda [28]2 years ago

6 0

$\cos \beta + \sin \beta \tan \beta \\\\= \cos \beta + (\tan \beta \cos \beta) \tan \beta \\\\=\cos \beta + \tan^2 \beta \cos \beta \\\\=\cos \beta( 1+ \tan^2 \beta)\\\\= \cos \beta \sec^2 \beta\\\\= \dfrac 1{\sec \beta} \times \sec^2 \beta \\\\= \sec \beta$

AleksandrR [38]2 years ago

3 0

Step-by-step explanation:

$\cos( \beta ) + \sin( \beta ) \tan( \beta ) = \sec( \beta )$

$\cos( \beta ) + \sin( \beta ) \frac{ \sin( \beta ) }{ \cos( \beta ) }$

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

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

$\frac{1}{ \cos( \beta ) }$

$= \sec( \beta )$

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Cindy is selling circus tickets. On the first day she sells 6 adult tickets and 5 children tickets for total of 112.50. On the s

Ivahew [28]

Answer: Children tickets cost 7.50

Adult tickets cost 12.50

Step-by-step explanation:

Let a represent adult tickets

Let c represent child tickets

On the first day she sells 6 adult tickets and 5 children tickets for total of 112.50. This can be written as:

6a + 5c = 112.50 ...... equation i

On the second day she sells 8 adult tickets and 4 children tickets for the total 130. This can be written as:

8a + 4c = 130 ....... equation ii

6a + 5c = 112.50 ...... equation i

8a + 4c = 130 ....... equation ii

Multiply equation i by 4

Multiply equation ii by 5

24a + 20c = 450 ........ equation iii

40a + 20c = 650 ......... equation iv

Subtract iii from iv

16a = 200

a = 200/16

a = 12.50

Adult tickets cost 12.50

From equation ii,

8a + 4c = 130

8(12.50) + 4c = 130

100 + 4c = 130

4c = 130 - 100

4c = 30

c = 30/4

c = 7.50

Children tickets cost 7.50

7 0

2 years ago

Is anyone able to figure this out, I can't do this

katrin [286]

Answer:

C. √2 - 1

Step-by-step explanation:

If we draw a square from the center of the large circle to the center of one of the small circles, we can see that the sides of the square are equal to the radius of the small circle (see attached diagram)

Let r = the radius of the small circle

Using Pythagoras' Theorem $a^2+b^2=c^2$

(where a and b are the legs, and c is the hypotenuse, of a right triangle)

to find the diagonal of the square:

$\implies r^2 + r^2 = c^2$

$\implies 2r^2 = c^2$

$\implies c=\sqrt{2r^2}$

So the diagonal of the square = $\sqrt{2r^2}$

We are told that the radius of the large circle is 1:

⇒ Diagonal of square + r = 1

$\implies \sqrt{2r^2}+r=1$

$\implies \sqrt{2r^2}=1-r$

$\implies 2r^2=(1-r)^2$

$\implies 2r^2=1-2r+r^2$

$\implies r^2+2r-1=0$

Using the quadratic formula to calculate r:

$\implies r=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\implies r=\dfrac{-2\pm\sqrt{2^2-4(1)(-1)}}{2(1)}$

$\implies r=\dfrac{-2\pm\sqrt{8}}{2}$

$\implies r=-1\pm\sqrt{2}$

As distance is positive, $r=-1+\sqrt{2}=\sqrt{2}-1$ only

5 0

2 years ago

Read 2 more answers

How do i find the sum or difference of this fraction

gregori [183]

1st use the Step! Please, Excuse, My, Dear, Aunt, Sally! Ehich also means (PEMDAS) Parenthesis (), Exponents ", Multiple ×, Division÷, Addition+, and Subtraction-. Than your going to solve! (3n+5)×9=n×7
27n+45=7n
27n-7n=-45
20n=-45
n=-45/20= -9/4

7 0

2 years ago

What’s the distance between (-3,-5) and (-2,-4)

joja [24]

Answer:

2.82843

Step-by-step explanation:

it was what i got