What are the steps for using a compass and straightedge to construct an quadrilateral triangle?

The following angle measures could form a triangle

qaws [65]

Answer:

They don't form a triangle because all triangles degrees have to add up to 180 if you add them all up its 190 so no they don't form a triangle.

Step-by-step explanation:

84+75+31=190

8 0

3 years ago

Xz=58 and yz=23.1 find xy

Pachacha [2.7K]

Step-by-step explanation:

$XZ = 58$

$YZ = 23.1$

$XY = XZ + YZ$

$XY =58 + 23.1$

$XY = 81 .1$

4 0

2 years ago

A thermometer gives a reading of 26 degrees Celsius. Use the formula c =5/9(f-32). Write the inverse function and use it to find

guajiro [1.7K]

Answer:

The answer to your question is F = 78.8

Step-by-step explanation:

Equation

$C = \frac{5}{9} (F - 32)$

Multiply both sides by 9

$9C = \frac{5(9)}{9} (F - 32)$

Simplify

$9C = 5(F - 32)$

Divide both sides by 5

$\frac{9}{5} C = \frac{5}{5} (F - 32)$

Simplify

$\frac{9}{5} C = F -32$

Add 32 in both sides

$\frac{9}{5} C + 32 = F - 32 + 32$

Simplify (Inverse)

$\frac{9}{5} C + 32 = F$

Substitute to find F

$\frac{9}{5} (26) + 32 = F$

Simplify

F = 46.8 + 32

Result

F = 78.8

6 0

3 years ago

Power Series Differential equation

KatRina [158]

The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for $y$

$\displaystyle\sum_{n\ge2}\bigg((n-3)(n-2)a_n+(n+3)(n+2)a_{n+3}\bigg)x^{n+1}+2a_2+(6a_0-6a_3)x+(6a_1-12a_4)x^2=0$

which indeed gives the recurrence you found,

$a_{n+3}=-\dfrac{n-3}{n+3}a_n$

but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that $a_2=0$ , and substituting this into the recurrence, you find that $a_2=a_5=a_8=\cdots=a_{3k-1}=0$ for all $k\ge1$ .

Next, the linear term tells you that $6a_0+6a_3=0$ , or $a_3=a_0$ .

Now, if $a_0$ is the first term in the sequence, then by the recurrence you have

$a_3=a_0$
$a_6=-\dfrac{3-3}{3+3}a_3=0$
$a_9=-\dfrac{6-3}{6+3}a_6=0$

and so on, such that $a_{3k}=0$ for all $k\ge2$ .

Finally, the quadratic term gives $6a_1-12a_4=0$ , or $a_4=\dfrac12a_1$ . Then by the recurrence,

$a_4=\dfrac12a_1$
$a_7=-\dfrac{4-3}{4+3}a_4=\dfrac{(-1)^1}2\dfrac17a_1$
$a_{10}=-\dfrac{7-3}{7+3}a_7=\dfrac{(-1)^2}2\dfrac4{10\times7}a_1$
$a_{13}=-\dfrac{10-3}{10+3}a_{10}=\dfrac{(-1)^3}2\dfrac{7\times4}{13\times10\times7}a_1$

and so on, such that

$a_{3k-2}=\dfrac{a_1}2\displaystyle\prod_{i=1}^{k-2}(-1)^{2i-1}\frac{3i-2}{3i+4}$

for all $k\ge2$ .

Now, the solution was proposed to be

$y=\displaystyle\sum_{n\ge0}a_nx^n$

so the general solution would be

$y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6+\cdots$
$y=a_0(1+x^3)+a_1\left(x+\dfrac12x^4-\dfrac1{14}x^7+\cdots\right)$
$y=a_0(1+x^3)+a_1\displaystyle\left(x+\sum_{n=2}^\infty\left(\prod_{i=1}^{n-2}(-1)^{2i-1}\frac{3i-2}{3i+4}\right)x^{3n-2}\right)$

4 0

3 years ago

In a large population, 82% of the households have cable tv. A simple random sample of 225 households is to be contacted and the

anzhelika [568]

Answer:

The mean of the sampling distribution of the sample proportions is 0.82 and the standard deviation is 0.0256.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .