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Pepsi [2]
3 years ago
12

Least comun multiple 6 and 8

Mathematics
2 answers:
ziro4ka [17]3 years ago
8 0

Answer:

one

Step-by-step explanation:

6                    8

1,6                 1,8

2,3                 2,4

konstantin123 [22]3 years ago
6 0

Answer:

The LCM of 8 and 6 is 24

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I have 2 triangle faces. i have 3 rectanglar faces. i have 6 vertices. i have 9 edges. what am i?
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Answer:

You are a rectangular prism.

Step-by-step explanation:

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3 years ago
Help please with algebra
KIM [24]

Answer:

c,d

Step-by-step explanation:

ab^{-3x}=a(b^{-3})^x=a(\frac{1}{b^3} )^x=a[(\frac{1}{b} )^3]^x\\=a(\frac{1}{b} )^{3x}

4 0
2 years ago
Please Help. I really don’t get this concept, if you could explain it in detail it would be very appreciated
Dafna11 [192]
<h3>Answer: 24/25</h3>

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

Explanation:

Sine is given to be negative, and so is tangent. This only happens in quadrant Q4

Recall that y = sin(theta), so if sin(theta) < 0, then we're below the x axis.

If tan(theta) < 0, then this means cos(theta) > 0

So we have y < 0 and x > 0 which places the angle somewhere in Q4.

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

Draw a right triangle as shown below in the attached image. We have AC = 25 and BC = 7. Use the pythagorean theorem to find that AB = 24

So this is what your steps may look like

a^2+b^2 = c^2

7^2+b^2 = 25^2

b^2+49 = 625

b^2 = 625-49

b^2 = 576

b = sqrt(576)

b = 24

So because AB = 24, we know that the cosine of the angle is adjacent/hypotenuse = 24/25

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

As an alternative, you could use the trig identity

sin^2(x) + cos^2(x) = 1

and plug in the given value of sine to solve for cosine. The cosine value result will be positive since we're in Q4.

So,

sin^2(x) + cos^2(x) = 1

(-7/25)^2 + cos^2(x) = 1

(49/625) + cos^2(x) = 1

cos^2(x) = 1 - (49/625)

cos^2(x) = (625/625) - (49/625)

cos^2(x) = (625-49)/625

cos^2(x) = 576/625

cos(x) = sqrt(576/625)

cos(x) = sqrt(576)/sqrt(625)

cos(x) = 24/25

This is effectively a rephrasing of the previous section since the pythagorean trig identity is more or less the pythagorean theorem (just in a trig form)

6 0
3 years ago
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