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Darya [45]
3 years ago
9

Given that v=√u*u- 2qs. Make q the subject of the equation. Find the value of q in the equation if v=8, u=12, and s=40​

Mathematics
2 answers:
densk [106]3 years ago
5 0

Above given answer is correct.

kifflom [539]3 years ago
3 0

Answer:

Step-by-step explanation:

v=√u*u- 2qs

  1. \sqrt{20} x 20 - 2q40 = 8
  2. 2\sqrt{5} x 20 - 2q40 = 8
  3. 40\sqrt{5} - 2q40 = 8
  4. - 2q40 = 8 - 40\sqrt{5}
  5. -80q = 8 - 40\sqrt{5}
  6. q = (8-40\sqrt{5} ) / -80 = \frac{-1 + 5\sqrt{5} }{10} = 1.018
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Identify whether the expressions are equivalent or not equivalent: 11(p + q) and 11p + (7q + 4q)
Stella [2.4K]

Answer:

Yes they are

Step-by-step explanation:

11(p+q) = 11p + 11q

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

11p + (7q+ 4q)

= 11p + ( q * (7+4))

= 11p + (q * 11)

= 11p+ 11q

See they are the same

3 0
3 years ago
7 . Your goal is to sell at least 50 boxes of cookies for your school fundraiser.
worty [1.4K]

Answer:

a. number of boxes sold ≥ 50

b. 26 + x ≥ 50

Step-by-step explanation:

8 0
3 years ago
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The domain of a function f(x) is the set of all integers from 0 to 10, and the range of f(x) is the set of all in
e-lub [12.9K]
I think that the correct answer is f(-1)=-8 and f(0)=-5
I hope this helps :D
6 0
3 years ago
Assume that​ women's heights are normally distributed with a mean given by mu equals 62.5 in​, and a standard deviation given by
kirza4 [7]

Answer: a) The probability is approximately = 0.5793

b) The probability is approximately=0.8810

Step-by-step explanation:

Given : Mean : \mu= 62.5\text{ in}

Standard deviation : \sigma = \text{2.5 in}

a) The formula for z -score :

z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}

Sample size = 1

For x= 63 in. ,

z=\dfrac{63-62.5}{\dfrac{2.5}{\sqrt{1}}}=0.2

The p-value = P(z

0.5792597\approx0.5793

Thus, the probability is approximately = 0.5793

b)  Sample size = 35

For x= 63 ,

z=\dfrac{63-62.5}{\dfrac{2.5}{\sqrt{35}}}\approx1.18

The p-value = P(z

= 0.8809999\approx0.8810

Thus , the probability is approximately=0.8810.

6 0
3 years ago
Can someone plz help me. How can you find the inequalities of 11/15 and 5/7. Next 5/9 and 7/13. Next 11/15 and 5/7. Lastly 5/9 a
LekaFEV [45]
To make this a little clearer, let's give the pairs of inequalities the same denominator:

<span>Question 1: 
</span>\frac{11}{15} ? \frac{5}{7}
First, apply the common denominator to the first fraction:
(\frac{11}{15})7 \\  \frac{11}{15} *  \frac{7}{7}  \\  \frac{11*7}{15*7}  \\  \frac{77}{105}
Do the same for the second:15( \frac{5}{7}) \\  \frac{5}{7}* \frac{15}{15} \\  \frac{5*15}{7*15}  \\  \frac{75}{105}
Nest, compare the two fractions:
\frac{77}{105} \ \textgreater \   \frac{75}{105}
Therefore:
\frac{11}{15} > \frac{5}{7}
<span>
Question Two:</span>
\frac{5}{9} ? \frac{7}{13}
Apply the common denominator to fraction one:
13( \frac{5}{9}) \\  \frac{5}{9} * \frac{13}{13}  \\  \frac{5*13}{9*13}  \\  \frac{65}{117}
Fraction two:
9(\frac{7}{13}) \\  \frac{7}{13} *  \frac{9}{9}  \\  \frac{7*9}{13*9}  \\  \frac{63}{117}
Evaluate:
\frac{65}{117} > \frac{63}{117}
Therefore:
<span>\frac{5}{9} > \frac{7}{13}
</span>
Hope this helps!
5 0
3 years ago
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