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1 year ago

Sin10+sin20+sin40+sin50=sin70+sin80.prove it

1 answer:

4 0

We are going to prove it like this:
Lets use the formula sinA+sinB=2sin(A+B/2)cos(A-B/2)
Now we are going to take the left side of equation
sin10+sin40+sin50+sin20
Arranging =(sin50+sin10)+(sin40+sin20)
Applying the above formula. =2sin(50+10/2)cos(50-10/2)+2sin(40+20/… =2sin(30)cos(20)+2sin(30)cos(10)
=2sin30{cos20+cos10}
Again using the formula
cosA+cosB= 2cos(A+B/2)cos(A-B/2)
=2sin30{2cos(20+10/2).cos(20-10/2)}
=2sin30{2cos(15).cos(5)}
=2(1/2){2cos15.cos5} as sin30=1/2
=2cos15.cos5
Taking right side of equation sin70+ sin80
Using the formula sinA+sinB
= 2sin(A+B/2)cos(A-B/2)
=2sin(70+80/2)cos(70-80/2)
=2sin75cos5
=2sin(90-15)cos5
=2cos15.cos5 <span>Hope this helps</span>

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Gigi earned $65 for 5 hours of gardening. She earned $90 for 9 hours of office work. Which statement correctly compares Gigi’s e

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She earned $3 more per hour for gardening than for office work.

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2 years ago

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A store sells 8 colors of balloons with at least 28 of each color. How many different combinations of 28 balloons can be chosen?

Len [333]

Answer:

(a) $Selection = 6724520$

(b) $At\ most\ 12 = 6553976$

(c) $At\ most\ 8 = 6066720$

(d) $At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

Step-by-step explanation:

Given

$Colors = 8$

$Balloons = 28$ --- at least

Solving (a): 28 combinations

From the question, we understand that; a combination of 28 is to be selected. Because the order is not important, we make use of combination.

Also, because repetition is allowed; different balloons of the same kind can be selected over and over again.

So:

$n => 28 + 8-1$ $= 35$

$r = 28$

$Selection = ^{35}^C_{28$

$Selection = \frac{35!}{(35 - 28)!28!}$

$Selection = \frac{35!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29*28!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29}{7!}$

$Selection = \frac{35*34*33*32*31*30*29}{7*6*5*4*3*2*1}$

$Selection = \frac{33891580800}{5040}$

$Selection = 6724520$

Solving (b): At most 12 red balloons

First, we calculate the ways of selecting at least 13 balloons

Out of the 28 balloons, there are 15 balloons remaining (i.e. 28 - 13)

So:

$n => 15 + 8 -1 = 22$

$r = 15$

Selection of at least 13 =

$At\ least\ 13 = ^{22}C_{15}$

$At\ least\ 13 = \frac{22!}{(22-15)!15!}$

$At\ least\ 13 = \frac{22!}{7!15!}$

$At\ least\ 13 = 170544$

Ways of selecting at most 12 =

$At\ most\ 12 = Total - At\ least\ 13$ --- Complement rule

$At\ most\ 12 = 6724520- 170544$

$At\ most\ 12 = 6553976$

Solving (c): At most 8 blue balloons

First, we calculate the ways of selecting at least 9 balloons

Out of the 28 balloons, there are 19 balloons remaining (i.e. 28 - 9)

So:

$n => 19+ 8 -1 = 26$

$r = 19$

Selection of at least 9 =

$At\ least\ 9 = ^{26}C_{19}$

$At\ least\ 9 = \frac{26!}{(26-19)!19!}$

$At\ least\ 9 = \frac{26!}{7!19!}$

$At\ least\ 9 = 657800$

Ways of selecting at most 8 =

$At\ most\ 8 = Total - At\ least\ 9$ --- Complement rule

$At\ most\ 8 = 6724520- 657800$

$At\ most\ 8 = 6066720$

Solving (d): 12 red and 8 blue balloons

First, we calculate the ways for selecting 13 red balloons and 9 blue balloons

Out of the 28 balloons, there are 6 balloons remaining (i.e. 28 - 13 - 9)

So:

$n =6+6-1 = 11$

$r = 6$

Selection =

$^{11}C_6 = \frac{11!}{(11-6)!6!}$

$^{11}C_6 = \frac{11!}{5!6!}$

$^{11}C_6 = 462$

Using inclusion/exclusion rule of two sets:

$Selection = At\ most\ 12 + At\ most\ 8 - (12\ red\ and\ 8\ blue)$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 170544+ 657800- 462$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = Total - Only\ 12\ red\ and\ only\ 8\ blue$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 6724520 - 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

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1 year ago

One factor of the function f(x) = x3 − 8x2 + 17x − 10 is (x − 5). Describe how to find the x-intercepts and the y-intercept of t

kozerog [31]

Answer:

x-intercepts: (1,0), (2,0), (5,0)

y-intercepts: (0,-10)

Step-by-step explanation:

One factor of the function $f(x) = x^3-8x^2 + 17x-10$ is $(x-5).$

Factor the polynomial in the following way:

$x^3-8x^2 + 17x-10\\ \\=x^3-5x^2-3x^2+15x+2x-10\\ \\=x^2(x-5)-3x(x-5)+2(x-5)\\ \\=(x-5)(x^2-3x+2)$

Now factor the quadratic in brackets:

$x^2-3x+2\\ \\=x^2-2x-x+2\\ \\=x(x-2)-(x-2)\\ \\=(x-2)(x-1)$

Hence,

$f(x)=(x-5)(x-2)(x-1)$

The x-intercepts are points at which y = 0, so

$(x-5)(x-2)(x-1)=0\\ \\x-5=0\ \text{or}\ x-2=0\ \text{or}\ x-1=0\\ \\x=5\ \text{or}\ x=2\ \text{or}\ x=1$

The y-intercepts are points at which x = 0, so

$y=0^3-8\cdot 0^2+17\cdot 0-10=-10$

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2 years ago

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After Luca took his dogs for a walk, he gave them 6 dog treats. When Luca's dad got home from work, he gave the dogs t more trea

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The equation would be 6+t=10 because they got 6 and then they got p which totaled 10

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2 years ago

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A walking trail is 1,580.76 feet long. A lake is located 70.62 feet away from the end of the trail. What is the total distance f

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Answer: $1651.38\ ft$

Step-by-step explanation:

Given

A walking trail is $1580.76\ ft$ long and a lake is located $70.62\ ft$ away from end of the trail

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2 years ago