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Lemur [1.5K]
3 years ago
9

Complete the pattern. 64, 32, 16, 8, _, _

Mathematics
2 answers:
Alja [10]3 years ago
5 0
4,2
Im pretty sure that’s right
HACTEHA [7]3 years ago
4 0

Answer:

64, 32, 16, 8, 4, 2

Step-by-step explanation:

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How to find an equation for a line through two given points?​
xz_007 [3.2K]

Answer:

The equation of the line is: y = 0.6x + 0.6

Step-by-step explanation:

Equation of a line:

The equation of a line has the following format:

y = mx + b

In which m is the slope and b is the y-intercept.

Two points:

We have these following two points in this exercise:

x = -6, y = -3, so (-6,-3)

x = 4, y = 3, so (4,3)

Finding the slope:

Given two points, the slope is given by the change in y divided by the change in x.

Change in y: 3 - (-3) = 3 + 3 = 6

Change in x: 4 - (-6) = 4 + 6 = 10

So

m = \frac{6}{10} = 0.6

Then

y = 0.6x + b

Finding b:

We replace one of the points in the equation to find b. I will use (4,3).

y = 0.6x + b

3 = 0.6*4 + b

2.4 + b = 3

b = 0.6

The equation of the line is: y = 0.6x + 0.6

4 0
2 years ago
279 is 33 1/2 of what amount
ycow [4]

Answer:

33% of 279 is equivalent to multiplying them: 33% × 279.

Step-by-step explanation:

4 0
3 years ago
Someone help please
Alla [95]

Answer:  Choice A

\tan(\alpha)*\cot^2(\alpha)\\\\

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

Explanation:

Recall that \tan(x) = \frac{\sin(x)}{\cos(x)} and \cot(x) = \frac{\cos(x)}{\sin(x)}. The connection between tangent and cotangent is simply involving the reciprocal

From this, we can say,

\tan(\alpha)*\cot^2(\alpha)\\\\\\\frac{\sin(\alpha)}{\cos(\alpha)}*\left(\frac{\cos(\alpha)}{\sin(\alpha)}\right)^2\\\\\\\frac{\sin(\alpha)}{\cos(\alpha)}*\frac{\cos^2(\alpha)}{\sin^2(\alpha)}\\\\\\\frac{\sin(\alpha)*\cos^2(\alpha)}{\cos(\alpha)*\sin^2(\alpha)}\\\\\\\frac{\cos^2(\alpha)}{\cos(\alpha)*\sin(\alpha)}\\\\\\\frac{\cos(\alpha)}{\sin(\alpha)}\\\\

In the second to last step, a pair of sine terms cancel. In the last step, a pair of cosine terms cancel.

All of this shows why \tan(\alpha)*\cot^2(\alpha)\\\\ is identical to \frac{\cos(\alpha)}{\sin(\alpha)}\\\\

Therefore, \tan(\alpha)*\cot^2(\alpha)=\frac{\cos(\alpha)}{\sin(\alpha)}\\\\ is an identity. In mathematics, an identity is when both sides are the same thing for any allowed input in the domain.

You can visually confirm that \tan(\alpha)*\cot^2(\alpha)\\\\ is the same as \frac{\cos(\alpha)}{\sin(\alpha)}\\\\ by graphing each function (use x instead of alpha). You should note that both curves use the exact same set of points to form them. In other words, one curve is perfectly on top of the other. I recommend making the curves different colors so you can distinguish them a bit better.

6 0
2 years ago
A right triangle has a leg length of 21 inches and a hypotenuse length of 29 inches. What is the length of the other leg?​
Shalnov [3]

Answer:

20

Step-by-step explanation:

Pythagorean Theorem: a^2 + b^2 = c^2

21^2 + b^2 = 29^2

441 + b^2 = 841

Subtract 441 on both sides to get b^2 = 400

Take the square root of both sides to get b = 20

4 0
3 years ago
calcule o montante de um capital de 4 000,00 empregado durante dois anos a seis meses área taxa de 1,5% a.m.
EastWind [94]
Can you plz translate all of this is English i dont understand ^-^-^-^
3 0
2 years ago
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