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adell [148]
3 years ago
7

I don't know how to solve it?

Mathematics
2 answers:
Law Incorporation [45]3 years ago
8 0
2/4+2/4=4/8 And 8×4=32 same as 8+8=16
16+16=32 its that easy
Grace [21]3 years ago
3 0
Just turn 5/6ths into x/32nds. For example I could turn 1/2 into 2/4. And 2/4 into 4/8.
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Can someone answer this Algebra question asap? thanks
Pavel [41]

Answer:

y = - 24

Step-by-step explanation:

Given that y varies directly with x then the equation relating them is

y = kx ← k is the constant of variation

To find k use the condition y = 12 when x = - 3

k = \frac{y}{x} = \frac{12}{-3} = - 4

y = - 4x ← equation of variation

When x = 6, then

y = - 4 × 6 = - 24

7 0
2 years ago
Does the ordered pair (−5,−5) satisfy the following system of equations?<br> {−3x−3y=125x+5y=−20
FrozenT [24]

Well, let's see. The problem gave you an ordered pair. In other words, you have an 'x' and a 'y' coordinate. All you need to do is put them into the equation.

Step-by-step explanation:

This means that instead of --

-3x - 3y = 125 + 5y = -20

We would have:

-3(-5)-3(-5) = 125(-5) + 5(-5) = -20

From here, you just simplify it into:

30 = -650 = -20

Since the values are not the same, the ANSWER is NO. The ordered pair does not satisfy the following system of equations.

3 0
3 years ago
Is 4 greater than 4.002?
elixir [45]
No because 4=4.00000 and there is that much more than 4.00000 so 4.002 is more
5 0
2 years ago
Read 2 more answers
1 + tanx / 1 + cotx =2
Lera25 [3.4K]

Answer:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

Step-by-step explanation:

Solve for x:

1 + cot(x) + tan(x) = 2

Multiply both sides of 1 + cot(x) + tan(x) = 2 by tan(x):

1 + tan(x) + tan^2(x) = 2 tan(x)

Subtract 2 tan(x) from both sides:

1 - tan(x) + tan^2(x) = 0

Subtract 1 from both sides:

tan^2(x) - tan(x) = -1

Add 1/4 to both sides:

1/4 - tan(x) + tan^2(x) = -3/4

Write the left hand side as a square:

(tan(x) - 1/2)^2 = -3/4

Take the square root of both sides:

tan(x) - 1/2 = (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

tan(x) = 1/2 + (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Take the inverse tangent of both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) = 1/2 - (i sqrt(3))/2

Take the inverse tangent of both sides:

Answer:  x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

4 0
3 years ago
Ming weighed a bag of soil for a science experiment. The weight was 2.003 kg. She then removed 0.520 kg of soil from the bag. Ho
natima [27]
Just subtract 2.003 and 0.520 which gives you 1.483 so there is 1.483 kilogram of soil remaining hope this helps!
7 0
3 years ago
Read 2 more answers
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