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3 years ago

NEED HELP ASAP!!!!!!!!

Mathematics

1 answer:

Zolol [24]3 years ago

5 0

Answer:

x = 2.4

y = 31

Step-by-step explanation:

Angle B = 25x

BC = 31

therefore , this is an equilateral triangle all sides and angles are equal,

25x =60

x = 2.4

BC = BA = AC

y= 31

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A set of raw paired sample data is given below. Convert this raw data into paired ranks, and calculate the value of the rs test

SOVA2 [1]

here is the data set for the complete question

x: 18 21 19 21 20 21

y; 2 14 5 6 18 18

Answer:

B. 0.652

Step-by-step explanation:

x y rank of x rank of y d d²

18 2 1 1 0 0

21 14 4 4 0 0

19 5 2 2 0 0

21 6 4 3 1 1

20 18 3 5.5 -2.5 6.25

21 18 4 5.5 -1.5 2.25

∑d² = 8.5

rs = 1 - 6[∑di² + ∑m(m²-1)]/n(n²-1)

= 1 - 6[8.5 +{3(3²-10/12 + 2(2² - 1)/12}]/6(6²-1)

= 1 - 0.348

= 0.652

therefore option b is the right answer.

5 0

3 years ago

1 5/16 + 1 5/16=_____

algol13

Answer:

2 10/16 or 2 5/8

Step-by-step explanation:

You can add the whole numbers first, so 1+1=2

Now add the remaining fractions: 5/16 +5/16=10/16.

2+10/16=2 10/16 or 2 5/8.

5 0

3 years ago

There are 6 diagonals that can be drawn from one vertex of an octagon, true or false.

FinnZ [79.3K]

It's False; an octagon has 8 vertices. When you remove the starting vertex and the two adjacent vertices we're left with 5 possible diagonals

6 0

3 years ago

What is the area of a regular nonagon (9 sides) with side length of 15 cm and apothem of 20.6 cm? Round the answer to the neares

Triss [41]

We know that

<span>The nine radii of a regular Nonagon divides into 9 congruent isosceles triangles
</span>therefore
[the area of <span>a regular nonagon]=9*[area of isosceles triangle]

</span>[area of isosceles triangle]=b*h/2------> 15*20.6/2----> 154.5 cm²
so
[the area of a regular nonagon]=9*[154.5]------> 1390.5 cm²

the answer is
1390.5 cm²

5 0

3 years ago

Find the general solution of the differential equation and check the result by differentiation. (Use C for the constant of integ

atroni [7]

Answer: $y=Ce^(^3^t^{^9}^)$

Step-by-step explanation:

Beginning with the first differential equation:

$\frac{dy}{dt} =27t^8y$

This differential equation is denoted as a separable differential equation due to us having the ability to separate the variables. Divide both sides by 'y' to get:

$\frac{1}{y} \frac{dy}{dt} =27t^8$

Multiply both sides by 'dt' to get:

$\frac{1}{y}dy =27t^8dt$

Integrate both sides. Both sides will produce an integration constant, but I will merge them together into a single integration constant on the right side:

$\int\limits {\frac{1}{y} } \, dy=\int\limits {27t^8} \, dt$

$ln(y)=27(\frac{1}{9} t^9)+C$

$ln(y)=3t^9+C$

We want to cancel the natural log in order to isolate our function 'y'. We can do this by using 'e' since it is the inverse of the natural log:

$e^l^n^(^y^)=e^(^3^t^{^9} ^+^C^)$

$y=e^(^3^t^{^9} ^+^C^)$

We can take out the 'C' of the exponential using a rule of exponents. Addition in an exponent can be broken up into a product of their bases:

$y=e^(^3^t^{^9}^)e^C$

The term e^C is just another constant, so with impunity, I can absorb everything into a single constant:

$y=Ce^(^3^t^{^9}^)$

To check the answer by differentiation, you require the chain rule. Differentiating an exponential gives back the exponential, but you must multiply by the derivative of the inside. We get:

$\frac{d}{dx} (y)=\frac{d}{dx}(Ce^(^3^t^{^9}^))$

$\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*\frac{d}{dx}(3t^9)$

$\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*27t^8$

Now check if the derivative equals the right side of the original differential equation:

$(Ce^(^3^t^{^9}^))*27t^8=27t^8*y(t)$

$Ce^(^3^t^{^9}^)*27t^8=27t^8*Ce^(^3^t^{^9}^)$

QED

I unfortunately do not have enough room for your second question. It is the exact same type of differential equation as the one solved above. The only difference is the fractional exponent, which would make the problem slightly more involved. If you ask your second question again on a different problem, I'd be glad to help you solve it.

7 0

2 years ago