All categories

2 years ago

Someone show equation form

Mathematics

1 answer:

Oliga [24]2 years ago

3 0

Answer/Step-by-step explanation:

$\begin{gathered}\boxed{\begin{array}{c|c}\boxed{\bf Form} &\boxed{\bf Equation} \\ \sf Slope\: intercept\:form &\sf y=mx+b \\ \sf Intercept\:form &\sf \dfrac{x}{a}+\dfrac{y}{b}=1\\ \sf Normal\:form &\sf xcos\omega+ysin\omega=p \\ \sf Two\: point\:form &\sf y-y_1=\left(\dfrac{y_2-y_1}{x_2-x_1}\right)(x-x_1)\\ \sf Point\:slope\:form &\sf y-y_1=m(x-x_1) \\ \sf Standard\:form &\sf Ax+By+C=0 \end{array}}\end{gathered}$

[RevyBreeze]

You might be interested in

Please someone help me...

Savatey [412]

use $a^2-b^2=(a+b)(a-b)$

to get $(\cos^3A-\sin^3A)(\cos^3A+\sin^3A)$

then use $a^3+b^3=(a+b)(a^2+b^2-ab)$

and $a^3-b^3=(a-b)(a^2+b^2+ab)$

also, $\sin^2\theta+\cos^2\theta=1$

to get $(\cos A-\sin A)(1+\sin A\cos A)(\cos A+ \sin A)(1-\sin A\cos A)$

then again use the first identity In both pairs, i.e.

$(\cos A-\sin A)(\cos A+ \sin A) \cdot (1+\sin A\cos A)(1-\sin A\cos A)$

to get $\cos 2A (1-\sin^2A\cos^2A)$

multiply and divide by 4 to get the RHS.

because, $\sin(2A)= 2\sin A \cos A$

squaring both sides, $\sin^2 (2A)=4\sin^2A\cos^2A$

3 0

3 years ago

Read 2 more answers

In a shipment of 20 packages, 7 packages are damaged. The packages are randomly inspected, one at a time, without replacement, u

horsena [70]

Answer:

$\large\boxed{\large\boxed{0.119}}$

Explanation:

You need to find the probability that exactly three of the first 11 inspected packages are damaged and the fourth is damaged too.

1. Start with the first 11 inspected packages:

a) The number of combinations in which 11 packages can be taken from the 20 available packages is given by the combinatory formula:

$C(m,n)=\dfrac{m!}{m!(m-n)!}$

$C(20,11)=\dfrac{20!}{11!\cdot(20-11)!}$

b) The number of combinations in which 3 damaged packages can be chossen from 7 damaged packages is:

$C(7,3)=\dfrac{7!}{3!\cdot(7-3)!}$

c) The number of cominations in which 8 good packages can be choosen from 13 good pacakes is:

$C(13,8)=\dfrac{13!}{8!\cdot(13-8)!}$

d) The number of cominations in which 3 damaged packages and 8 good packages are chosen in the first 11 selections is:

$C(7,3)\times C(13,8)$

e) The probability is the number of favorable outcomes divided by the number of possible outcomes, then that is:

$\dfrac{C(7,3)\times C(13,8)}{C(20,11)}$

Subsituting:

$\dfrac{\dfrac{7!}{3!\cdot(7-3)!}\times \dfrac{13!}{8!\cdot(13-8)!}}{\dfrac{20!}{11!\cdot(20-11)!}}$

$=\dfrac{\dfrac{7!}{3!\cdot 4!}\times \dfrac{13!}{8!\cdot 5!}}{\dfrac{20!}{11!\cdot 9!}}=0.26818885$

2. The 12th package

The probability 12th package is damaged too is 7 - 3 = 4, out of 20 - 11 = 9:

3. Finally

The probability that exactly 12 packages are inspected to find exactly 4 damaged packages is the product of the two calculated probabilities:

$0.26818885\times 4/9=0.119$

6 0

3 years ago

Simplify the expression (x12)3

Grace [21]

I think its just 36x

3 0

4 years ago

A student prepares for an exam by studying a list of 10 problems. She can solve 8 of them. For the exam, the instructor selects

lianna [129]

Answer:

Probability to solve 7 problem will be equal to $\frac{1}{15}$

Step-by-step explanation:

It is given total number of question = 10

She can solve 8 of them from 10 questions

We have to find the probability that she can solve 7 question

Probability will be equal to ratio of favorable outcome to total sample space

Favorable outcome will be equal to $^8c_7$

Total sample space $=^{10}c_7$

So probability will be equal to $P=\frac{^8C_7}{^{10}C_7}=\frac{6}{90}=\frac{1}{15}$

So probability to solve 7 problem will be equal to $\frac{1}{15}$

5 0

4 years ago

HELP PICTURE IS SHOWN

Ahat [919]

Its 108 because the triangle is acute so all angle are less than or equal to 90 and i did 54+54+72=180 ant the measurement of the exterior angle is 108

5 0

4 years ago