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3 years ago

How do you rearrange x = 3g + 2 to make g the subject

Mathematics

1 answer:

Kazeer [188]3 years ago

4 0

X=3g+2
subtract 2 on both sides
x-2=3g
divide by 3 on both sides
x/3-2/3=g

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Whats the difference between 7 7/8 - 3 1/4

faust18 [17]

7 7/8 - 3 1/4

7 7/8 = 6.125

3 1/4 = 0.75

so 6.125 - 0.75

= 5.375

as a mixed fraction the answer is 5 3/8

4 0

3 years ago

Absolute value is always stated as a negative number,<br> True or false

blsea [12.9K]

Answer:

false

Step-by-step explanation:

hope it will be helpful

8 0

2 years ago

Can someone help meeee

GarryVolchara [31]

Answer:

1. 0

2. undefined

3. -3/2

Step-by-step explanation:

y=-3/2. This value is constant and the x does not change the value at all. so the coordinate are going to be (x, -3/2) so if we take any 2 values you're going to get (-3/2 - (-3/2)) / (x2 - x1) = (-3/2 + 3/2) / (x2 - x1) = (0) / (x2 - x1) which is always going to be 0.

x=-3/2. This value is also constant except the x-value does not vary at all so x2 and x1 are always going to be equal which will result in a 0 as the denominator thus making the slope undefined

y=-3/2x. it's just -3/2 given the slope-intercept form y=mx+b where m is the slope

4 0

1 year ago

Read 2 more answers

(cosxtanx-tanx+2cosx-2)/tanx+2=cosx-1<br> can someone show me how to prove this?

uysha [10]

Answer:

See detail below.

Step-by-step explanation:

A word of caution before getting to the actual problem: I believe there is an important set of brackets missing in the original post. The expression on the left hand side should be:

(cosxtanx-tanx+2cosx-2)/(tanx+2)

Without the brackets, it is left unclear whether the denominator is just tanx or tanx+2. I recommend to use brackets wherever any doubt could arise.

Now to the actual problem: \we can make the following transformations on the left hand side:

$\frac{\cos x \tan x + 2\cos x -2}{\tan x +2}=\frac{\cos x \frac{\sin x}{\cos x} + 2\cos x -2}{\frac{\sin x +2\sin x}{\cos x} }=\\=\frac{\cos x \sin x - \sin x + 2\cos^2 x - 2\cos x}{\sin x + 2\cos x}=\\=\frac{\cos x \sin x +2\cos^2 x }{\sin x + 2\cos x}-1=\\=\cos x -1$

which is shown to be the same as the right hand side, which was to be shown.

7 0

2 years ago

Pls give an explanation w your answer

Shalnov [3]

$a^{\frac{1}{n}}=\sqrt[n]{a}\\\\\left(a^n\right)^m=a^{n\cdot m}\\\\a^n\cdot a^m=a^{n+m}$
therefore
$64^\frac{1}{4}=\left(2^6\right)^\frac{1}{4}=2^{6\cdot\frac{1}{4}}\\\\=2^\frac{6}{4}=2^{1\frac{2}{4}}=2^{1+\frac{2}{4}}=2\cdot2^\frac{2}{4}\\\\=2\sqrt[4]{2^2}=2\sqrt[4]4$
Other method:
$64^\frac{1}{4}=\sqrt[4]{64}=\sqrt[4]{16\cdot4}=\sqrt[4]{16}\cdot\sqrt[4]4=2\sqrt[4]4$

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4 0

2 years ago

Read 2 more answers