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Simora [160]
4 years ago
8

What is the value of y?

Mathematics
1 answer:
Aleks04 [339]4 years ago
3 0
To solve for the value of y, all you have to do is add up all the expressions and values together and make it equal to 180 degrees.

Y + (Y - 12) + 56 = 180
2Y - 12 + 56 = 180
2Y + 44 = 180
Y = 68 degrees.
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P=m/1+rt solve for t
blagie [28]
I assume you mean:

P=m/(1+rt)  multiply both sides by (1+rt)

P(1+rt)=m  perform indicated multiplication on left side

P+Prt=m  subtract P from both sides

Prt=m-P  divide both sides by Pr

t=(m-P)/(Pr)
5 0
3 years ago
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1. The full Renegade dance Jalaiah created is 56 seconds long. For a 15-second-long TikTok, what percent of Renegade is used?​
VikaD [51]

Answer:

3.7

Step-by-step explanation:

You have to divide

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3 years ago
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I don't understand how to begin this problem
Ronch [10]
I would first add 5 and 2 together, then multiply that by 3. Finally you would subtract the 1. After solving for what is inside the parentheses you would multiply it by 2 for your final answer.

2•(3(5+2)-1)
2•(3(7)-1)
2•(21-1)
2•(20)
40
7 0
3 years ago
<img src="https://tex.z-dn.net/?f=5%20%5Csqrt%7B15%7D%20" id="TexFormula1" title="5 \sqrt{15} " alt="5 \sqrt{15} " align="absmid
PSYCHO15rus [73]
The answer is 1.71......
5 0
3 years ago
Consider the following. f(x) = x5 − x3 + 6, −1 ≤ x ≤ 1 (a) Use a graph to find the absolute maximum and minimum values of the fu
kykrilka [37]

ANSWER

See below

EXPLANATION

Part a)

The given function is

f(x) =  {x}^{5}  -  {x}^{3}  + 6

From the graph, we can observe that, the absolute maximum occurs at (-0.7746,6.1859) and the absolute minimum occurs at (0.7746,5.8141).

b) Using calculus, we find the first derivative of the given function.

f'(x) = 5 {x}^{4} - 3 {x}^{2}

At turning point f'(x)=0.

5 {x}^{4} - 3 {x}^{2}  = 0

This implies that,

{x}^{2} (5 {x}^{2}  - 3) = 0

{x}^{2}  = 0 \: or \: 5 {x}^{2}  - 3 = 0

x =  - \frac{ \sqrt{15} }{5}   \: or \: x = 0 \:  \: or \: x =\frac{ \sqrt{15} }{5}

We plug this values into the original function to obtain the y-values of the turning points

(   -  \frac{ \sqrt{15} }{5}  , \frac{1}{125} ( 6 \sqrt{15}  +750)) \:and \:  (0, - 6) \: and\: (   \frac{ \sqrt{15} }{5}  , \frac{1}{125} ( - 6 \sqrt{15}  +750))

We now use the second derivative test to determine the absolute maximum minimum on the interval [-1,1]

f''(x) = 20 {x}^{3}  - 6x

f''( -  \frac{ \sqrt{15} }{5} ) \:   <  \: 0

Hence

(   -  \frac{ \sqrt{15} }{5}  , \frac{1}{125} ( 6 \sqrt{15}  + 750))

is a maximum point.

f''( \frac{ \sqrt{15} }{5} ) \:    >  \: 0

Hence

(     \frac{ \sqrt{15} }{5}  , \frac{1}{125} (- 6 \sqrt{15}  + 750))

is a minimum point.

f''(0) \: =\: 0

Hence (0,-6) is a point of inflexion

4 0
3 years ago
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