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4 years ago

(6.7 x 10 5) – (2.3 x 102) =

Mathematics

2 answers:

Furkat [3]4 years ago

5 0

Answer:

6.7 x 105 = 703.5

2.3 x 102 = 234.6

703.5 - 234.6 = 468.9

Step-by-step explanation:

Bingel [31]4 years ago

4 0

-Answer:-10

Step-by-step explanation:

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densk [106]

Answer:The first one 18in.2

Step-by-step explanation:

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3 years ago

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Compare the following decimals 0.29 0.10 0.1 5.07 5.6

Kay [80]

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0.10, 0.1, 0.29, 5.07, 5.6

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8 0

4 years ago

135% as a fraction or a mixed number fully simplified please 50 POINTS + BRAINILEST

Valentin [98]

Answer:

27/20

Step-by-step explanation:

135% = 135/100 = (27/20 × 5) = 27/20 = 1 7/20 I think

4 0

3 years ago

Twin brothers, Billy and Bobby, can mow their grandparent’s lawn together in 63 minutes. Billy could mow the lawn by himself in

Pepsi [2]

Answer:

<h2>140 minutes</h2>

Step-by-step explanation:

this problem focuses on the combined work rate

both can finish the work under 63 minute

let bobby mow the lawn in x minute

Billy will then mow the lawn in x-25 minutes

Let the completed job = 1 (a mowed lawn)

we can now apply the combined work formula for the two boys as

$\frac{63}{x}+\frac{63}{x-25}=1$

multiplying through by x(x-25 )we have

$63(x-25)+63x= x(x-25)$

open bracket we have

$63x-1575+63x= x^2-25x$

rearranging and collecting like terms we have

$0=x^2-25x-63x-63x+1575 \\\\0=x^2-151x+1575 \\\\x^2-151x+1575=0$

we can now use the quadractic formula

$x=\frac{-b\frac{+}{} \sqrt{b^2-4ac} }{2a}$

a= 1

b= -151

c= 1575

$x=\frac{-(-151)\frac{+}{} \sqrt{151^2-4*1*1575} }{2*1}$

$x=\frac{-(-151)\frac{+}{} \sqrt{22801-6300} }{2} \\\\x=\frac{-(-151)\frac{+}{} \sqrt{16501} }{2} \\\\x=\frac{-(-151)\frac{+}{} 128.45}{2} \\\\$

$x=\frac{-(-151)+128.45}{2} \\\\\x=\frac{279.45}{2} \\\\x=139.72 \\\\\\x=\frac{-(-151)-128.45}{2}\\\\x=\frac{151-128.45}{2}\\\\x=\frac{22.55}{2}\\\\x=11.27$

the first answer x= 139.72 approximately 140 minutes mins is correct for Bobby's time

let us check

$\frac{63}{140}+\frac{63}{140-25}=1\\\\\\0.45+\frac{63}{115} =1\\\\0.45+0.5478=1\\0.997 =1$

we can see that the solution approximates to 1

8 0

4 years ago

Prove it please <br>answer only if you know

deff fn [24]

Part (c)

We'll use this identity

$\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)\\\\$

to say

$\sin(A+45) = \sin(A)\cos(45) + \cos(A)\sin(45)\\\\\sin(A+45) = \sin(A)\frac{\sqrt{2}}{2} + \cos(A)\frac{\sqrt{2}}{2}\\\\\sin(A+45) = \frac{\sqrt{2}}{2}(\sin(A)+\cos(A))\\\\$

Similarly,

$\sin(A-45) = \sin(A + (-45))\\\\\sin(A-45) = \sin(A)\cos(-45) + \cos(A)\sin(-45)\\\\\sin(A-45) = \sin(A)\cos(45) - \cos(A)\sin(45)\\\\\sin(A-45) = \sin(A)\frac{\sqrt{2}}{2} - \cos(A)\frac{\sqrt{2}}{2}\\\\\sin(A-45) = \frac{\sqrt{2}}{2}(\sin(A)-\cos(A))\\\\$

-------------------------

The key takeaways here are that

$\sin(A+45) = \frac{\sqrt{2}}{2}(\sin(A)+\cos(A))\\\\\sin(A-45) = \frac{\sqrt{2}}{2}(\sin(A)-\cos(A))\\\\$

Therefore,

$2\sin(A+45)*\sin(A-45) = 2*\frac{\sqrt{2}}{2}(\sin(A)+\cos(A))*\frac{\sqrt{2}}{2}(\sin(A)-\cos(A))\\\\2\sin(A+45)*\sin(A-45) = 2*\left(\frac{\sqrt{2}}{2}\right)^2\left(\sin^2(A)-\cos^2(A)\right)\\\\2\sin(A+45)*\sin(A-45) = 2*\frac{2}{4}\left(\sin^2(A)-\cos^2(A)\right)\\\\2\sin(A+45)*\sin(A-45) = \sin^2(A)-\cos^2(A)\\\\$

The identity is confirmed.

==========================================================

Part (d)

$\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)\\\\\sin(45+A) = \sin(45)\cos(A) + \cos(45)\sin(A)\\\\\sin(45+A) = \frac{\sqrt{2}}{2}\cos(A) + \frac{\sqrt{2}}{2}\sin(A)\\\\\sin(45+A) = \frac{\sqrt{2}}{2}(\cos(A)+\sin(A))\\\\$

Similarly,

$\sin(45-A) = \sin(45 + (-A))\\\\\sin(45-A) = \sin(45)\cos(-A) + \cos(45)\sin(-A)\\\\\sin(45-A) = \sin(45)\cos(A) - \cos(45)\sin(A)\\\\\sin(45-A) = \frac{\sqrt{2}}{2}\cos(A) - \frac{\sqrt{2}}{2}\sin(A)\\\\\sin(45-A) = \frac{\sqrt{2}}{2}(\cos(A)-\sin(A))\\\\$

-----------------

We'll square each equation

$\sin(45+A) = \frac{\sqrt{2}}{2}(\cos(A)+\sin(A))\\\\\sin^2(45+A) = \left(\frac{\sqrt{2}}{2}(\cos(A)+\sin(A))\right)^2\\\\\sin^2(45+A) = \frac{1}{2}\left(\cos^2(A)+2\sin(A)\cos(A)+\sin^2(A)\right)\\\\\sin^2(45+A) = \frac{1}{2}\cos^2(A)+\frac{1}{2}*2\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\\sin^2(45+A) = \frac{1}{2}\cos^2(A)+\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\$

and

$\sin(45-A) = \frac{\sqrt{2}}{2}(\cos(A)-\sin(A))\\\\\sin^2(45-A) = \left(\frac{\sqrt{2}}{2}(\cos(A)-\sin(A))\right)^2\\\\\sin^2(45-A) = \frac{1}{2}\left(\cos^2(A)-2\sin(A)\cos(A)+\sin^2(A)\right)\\\\\sin^2(45-A) = \frac{1}{2}\cos^2(A)-\frac{1}{2}*2\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\\sin^2(45-A) = \frac{1}{2}\cos^2(A)-\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\$

--------------------

Let's compare the results we got.

$\sin^2(45+A) = \frac{1}{2}\cos^2(A)+\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\\sin^2(45-A) = \frac{1}{2}\cos^2(A)-\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\$

Now if we add the terms straight down, we end up with $\sin^2(45+A)+\sin^2(45-A)$ on the left side

As for the right side, the sin(A)cos(A) terms cancel out since they add to 0.

Also note how $\frac{1}{2}\cos^2(A)+\frac{1}{2}\cos^2(A) = \cos^2(A)$ and similarly for the sin^2 terms as well.

The right hand side becomes $\cos^2(A)+\sin^2(A)$ but that's always equal to 1 (pythagorean trig identity)

This confirms that $\sin^2(45+A)+\sin^2(45-A) = 1$ is an identity

4 0

3 years ago