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agasfer [191]
3 years ago
15

Will a negative number become positive when exponent is positive fraction

Mathematics
1 answer:
-Dominant- [34]3 years ago
5 0
Yes, because a negative number multiply by itself it is positive. 
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Please help me and answer with work! Due tomorrow!​
ELEN [110]

Answer:

Step-by-step explanation:

a.

The angles given are 62° and 90°.

We know sum of angles in any triangle is 180° so the third angle must be 180-90-62 = 28°

We can find the sides if we know trigonometric functions

sin28° = opp. side/ hypothenuse = CA/BA = 8/BA

BA = 8/sin 28° ≈ 17

tan62°= opp.side/adj.side= BC/AC = BC/8

BC = 8 · tan62° ≈ 15

b.

We are given 2 sides 8.5 and 6.5 and that one angle is right, so the triangle is a right triangle, therefore we can apply Pythagorean Theorem to find the third side.

6.5² +FD² = 8.5², subtract 6.5² from both sides

         FD² = 8.5²- 6.5², square and combine like terms

         FD² = 30, square-root both sides

          FD = √30

          FD ≈ 5.47722, round to the nearest tenth

          FD ≈ 5.5

We can find the angles if we know trigonometric functions.

sin ∡D = FE/DE = 6.5/8.5

     ∡D = sin^-1 (6.5/8.5)

    ∡D ≈ 49.9°

cos ∡E = FE/DE = 6.5/8.5

      ∡E = cos^-1 (6.5/8.5)

      ∡E ≈ 40.1°

6 0
2 years ago
Please help!! 2 - 7 when x = - 1
mylen [45]
2-7 when x=-1?
2-7=-5?
What’s x=-1 information for?
4 0
3 years ago
Use​ l'Hôpital's Rule to find the following limit. ModifyingBelow lim With x right arrow 0StartFraction 3 sine (x )minus 3 x Ove
steposvetlana [31]

Answer:

\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=-\frac{1}{14}

Step-by-step explanation:

The limit is:

\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{0}{0}

so, you have an indeterminate result. By using the l'Hôpital's rule you have:

\lim_{x \to 0} \frac{a(x)}{b(x)}= \lim_{x \to 0} \frac{a'(x)}{b'(x)}

by replacing, and applying repeatedly you obtain:

\lim_{x \to 0} \frac{3sinx-3x}{7x^3}= \lim_{x \to 0}\frac{3cosx-3}{21x^2}= \lim_{x \to 0}\frac{-3sinx}{42x}= \lim_{x \to 0}\frac{-3cosx}{42}\\\\ \lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{-3cos0}{42}=-\frac{1}{14}

hence, the limit of the function is -1/14

8 0
3 years ago
There it is pls help me with this
andriy [413]
Where is the question? we can’t answer it without it
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3 years ago
ASAP 30 POINTS PLEASE HELP
dimulka [17.4K]

Answer:

the total cost of producing 6 widgets is $231

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