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Nataly_w [17]
2 years ago
7

Can anyone solve it.??​

Mathematics
2 answers:
bulgar [2K]2 years ago
3 0
The answer will be f √7 / 3 + 5 √7 / 14.
Kruka [31]2 years ago
3 0

\dfrac{\sqrt7}{3}+\dfrac{5}{2\sqrt7}=\dfrac{\sqrt7}{3}+\dfrac{5\sqrt7}{14}=\dfrac{14\sqrt7}{42}+\dfrac{15\sqrt7}{42}=\dfrac{29\sqrt7}{42}

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Factor the trigonometric equation.<br> cos2x-cosx
prisoha [69]

Cos(2x) = cos^2(x) - sin^2(x) - cos(x)

but sin^2(x) = 1 - cos^2(x)

cos(2x) - cos(x) = cos^2(x) - (1 - cos^2(x) ) - cos(x)

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

cos(2x) - cos(x) = 2cos^2(x) - 1 - cos(x)

cos(2x) - cos(x) = (2cos(x) + 1)(cos(x) - 1)

I think this is what you have asked for.

4 0
3 years ago
One small tank takes twice as long to fill a tank as a larger pipe does. If it takes the two pipes 40 minutes to fill the tank w
Gala2k [10]
Um... it'll take each pipe 80 mins, bcz if together they take that long, then to find the pipes time separately, you just multiply it by two to show it doing the work the other pipe did as well. Is that what you wanted??

8 0
3 years ago
Jane is 5 times older than her sister. In 3 years, Jane’s sister will be 1/4 her age. Find the ages now
Korolek [52]

The present age of Jane is 45 years old and present age of her sister is 9 years old

<em><u>Solution:</u></em>

Let the present age of Jane be "x"

Let the present age of her sister be "y"

<em><u>Jane is 5 times older than her sister</u></em>

present age of Jane = 5(present age of her sister)

x = 5y ---------- eqn 1

<em><u>In 3 years, Jane’s sister will be 1/4 her age</u></em>

Age of sister after 3 years = 3 + y

Age of jane after 3 years = 3 + x

Age of sister after 3 years = 1/4(age of jane after 3 years)

3 + y = \frac{1}{4}(3 + x)

Substitute eqn 1 in above equation

3 + y = \frac{1}{4}(3 + 5y)\\\\12 + 4y = 3 + 5y\\\\5y - 4y = 12 - 3\\\\y = 9

Substitute y = 9 in eqn 1

x = 5(9)

x = 45

Thus present age of Jane is 45 years old and present age of her sister is 9 years old

3 0
3 years ago
Help on this question explain my answer please!
Tresset [83]
0.12 x 70 = 8.4

12% of 70 is 8.4
5 0
3 years ago
Suppose babies born after a gestation period of 32 to 35 weeks have a mean weight of 2500 grams and a standard deviation of 700
const2013 [10]

Answer:

Z-score for 34-week baby = 0.643

Z-score for 41-week baby = 1.154

Baby weight of 41-week is more than the baby weight of 34-week in the gestation period.

Step-by-step explanation:

Given - Suppose babies born after a gestation period of 32 to 35 weeks have a mean weight of 2500 grams and a standard deviation of 700 grams while babies born after a gestation period of 40 weeks have a mean weight of 3100 grams and a standard deviation of 390 grams. If a 34-week gestation period baby weighs 2950 grams and a 41-week gestation period baby weighs 3550 grams

To find - Find the corresponding z-scores. Which baby weighs more relative to the gestation period.

Proof -

Given that,

In between period of 32 to 35 weeks

Mean = 2500

Standard deviation = 700

In between after a period of 40 weeks

Mean = 3100

Standard deviation = 390

Now,

For a 34-week baby,

X = 2950

For a 41-week baby,

X = 3550

Now,

Z-score = (X - mean) / Standard deviation

Now,

For a 34-week baby,

Z - score = (2950 - 2500) / 700 = 0.643

For a 41-week baby,

Z-score = (3550 - 3100) / 390 = 1.154

∴ we get

Z-score for 34-week baby = 0.643

Z-score for 41-week baby = 1.154

As 1.154 > 0.643

So,

Baby weight of 41-week is more than baby weight of 34-week in the gestation period.

4 0
3 years ago
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