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3 years ago

The number 360 is increased by 25%. The result is then decreased by 50%. What is the final number?

1 answer:

3 0

So if 360 is increased by 25%, you would end up with 450 (360 x 1.25), than if the result, 450 is decreased by 50% ( 450 x .50 = 225) (450 - 225 = 225) you would end up with 225.

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andy is 5 kilogram less than twice his brother's. together they weigh 100 kilograms what are they weights

Minchanka [31]

<span>Let x be the brother's weight.
Then 2x-5 is Andy's weight.
x+(2x-5)=100
3x-5=100
3x=105</span>

6 0

3 years ago

Read 2 more answers

Which of the following is a solution to the system of equations? y= 3x+24<br> y=10-0.5x

Sergeeva-Olga [200]

Answer:

x = -4

y = 12

I hope it helps you

7 0

3 years ago

PLEASE HELP ASAP! URGENT!

Nonamiya [84]

Answer:

It would be a negative number, no matter what negative integer you put it as.

Step-by-step explanation:

For example, lets substitute -2

f(-2) = -8-4-1 = -13; a negative number

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3 years ago

Prove that sin3a-cos3a/sina+cosa=2sin2a-1

Sloan [31]

Answer:

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

Step-by-step explanation:

we are given

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

we can simplify left side and make it equal to right side

we can use trig identity

$sin(3a)=3sin(a)-4sin^3(a)$

$cos(3a)=4cos^3(a)-3cos(a)$

now, we can plug values

$\frac{(3sin(a)-4sin^3(a))-(4cos^3(a)-3cos(a))}{sin(a)+cos(a)}$

now, we can simplify

$\frac{3sin(a)-4sin^3(a)-4cos^3(a)+3cos(a)}{sin(a)+cos(a)}$

$\frac{3sin(a)+3cos(a)-4sin^3(a)-4cos^3(a)}{sin(a)+cos(a)}$

$\frac{3(sin(a)+cos(a))-4(sin^3(a)+cos^3(a))}{sin(a)+cos(a)}$

now, we can factor it

$\frac{3(sin(a)+cos(a))-4(sin(a)+cos(a))(sin^2(a)+cos^2(a)-sin(a)cos(a)}{sin(a)+cos(a)}$

$\frac{(sin(a)+cos(a))[3-4(sin^2(a)+cos^2(a)-sin(a)cos(a)]}{sin(a)+cos(a)}$

we can use trig identity

$sin^2(a)+cos^2(a)=1$

$\frac{(sin(a)+cos(a))[3-4(1-sin(a)cos(a)]}{sin(a)+cos(a)}$

we can cancel terms

$=3-4(1-sin(a)cos(a))$

now, we can simplify it further

$=3-4+4sin(a)cos(a))$

$=-1+4sin(a)cos(a))$

$=4sin(a)cos(a)-1$

$=2\times 2sin(a)cos(a)-1$

now, we can use trig identity

$2sin(a)cos(a)=sin(2a)$

we can replace it

$=2sin(2a)-1$

so,

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

7 0

3 years ago

Read 2 more answers

How do you solve this whole thing please?

NemiM [27]

Answer:

Step-by-step explanation:

Let's look at the first two, and hopefully you'll be able to figure the rest out:

1. Answer below

$d = rt$

The problem asks us to solve for $r$ , so that means we need to get $r$ on one side of the equation by itself. To do so, we will need to divide both sides of the equation by $t$ :

$\frac{d}{t} = \frac{rt}{t}$

$\frac{d}{t} = r$

2. Answer below

$B = T - Lc$

The problem asks us to solve for $T$ , so that means we need to get $T$ on one side of the equation by itself. To do so, we will need to add $Lc$ to both sides of equation:

$B + Lc = T - Lc + Lc$

$B + Lc = T$

7 0

3 years ago