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yawa3891 [41]
3 years ago
11

Find the 31st term of the following sequence. 9, 15, 21, ...

Mathematics
1 answer:
Marrrta [24]3 years ago
8 0
The 31st term of this sequence is 189.
 9, 15, 21, 27, 33 (5), 39, 45, 51, 57, 63 (10), 69, 75, 81, 87, 93 (15), 99, 105, 111, 117, 123 (20), 129, 135, 141, 147, 153 (25), 159, 165, 171, 177, 183 (30), 189.
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Simplify the expression−6v + 3v
Maurinko [17]

Answer:

-3v

Step-by-step explanation:

-6v is a negative when 3v is positive

when you add 3v to -6v it equals -3v since 6 is greater

3 0
3 years ago
Read 2 more answers
I need y’all to help me asap!!!!!!
timofeeve [1]

Answer:

-1

Step-by-step explanation:

f(0) = -5^0 = -1

8 0
3 years ago
Rewrite this polynomial in STANDARD FORM:<br>5x-10x^2+8x^3​
Ainat [17]

Answer:

8 {x}^{3}   -  10 {x}^{2}  + 5x

Step-by-step explanation:

5x - 10 {x}^{2} + 8 {x}^{3}   \\   \red{ \bold{= 8 {x}^{3}   -  10 {x}^{2}  + 5x}} \\ is \: in \: the \: standard \: form \\

4 0
3 years ago
1. Which of the following does NOT represent a whole number? 9 6.0 12/4 -12/3​
Ivahew [28]

the answer is -12/3 cause it's negative

5 0
3 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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