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2 years ago

AV Can the HL Congruence Property be used? Yes O No O Yes O No

Mathematics

1 answer:

Svetradugi [14.3K]2 years ago

6 0

Answer:

yes! I hope this helps and I hope it's correct I'm and pretty sure it is.

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Given that x and y vary inversely and that y is 8 when x is 2, what is the constant of variation, k?

Shtirlitz [24]

Answer:

Step-by-step explanation:

the equation is x = k / y

therefore k = x* y

k = 2*8

k= 16

4 0

3 years ago

A libary has 2000 books. there are 3 times as many non fiction books as fiction books. write and solve a system of equations to

Andru [333]

A: number of nonfiction books
b: number of fiction books
a+b=2000
a=3b
b=500
a=1500

4 0

3 years ago

A. How many ways can the letters of the word ALGORITHM be arranged in a row?

Tpy6a [65]

Answer:

a). The number if ways it can be arranged in a row= 362880 ways

b). 40320 ways

c). 5040 ways

Step-by-step explanation:

There are 9 alphabets in the word ALGORITHM

The number if ways it can be arranged in a row= 9!

The number if ways it can be arranged in a row= 9*8*7*6*5*4*3*2*1

The number if ways it can be arranged in a row= 362880 ways

if A and L must remain together (in order) as a unit, then we take it as 8 alphabets

= 8!

= 8*7*6*5*4*3*2*1

= 40320 ways

if the letters GOR must remain together (in order) as a unit, then we have 7 alphabets units remaining

= 7!

= 7*6*5*4*3*2*1

= 5040 ways

8 0

3 years ago

g Given the system of equations: 15 c1 – 5 c2 – c3 = 2400 - 5 c1 + 18 c2 – 6 c3 = 3500 - 4 c1 - c2 + 12 c3 = 4000 (a) Calculate

denpristay [2]

Answer

(a)

$A^{-1} = \frac{1}{3493} \begin{pmatrix} 210 && -59 && -12 \\ 84 && 176 && 95 \\ 77 && -5 && 295 \end{pmatrix}$

(b)

$A^{-1} b = \begin{pmatrix} \frac{500}{7} \\\\ \frac{2400}{7} \\\\ \frac{2700}{7} \end{pmatrix}$

Step-by-step explanation:

Remember that when you want to solve a problem like this, you express the equation as following

$Ax = b$

So, if you know the inverse of $A$ then

$x = A^{-1} b$

For this case

$A = \begin{pmatrix} 15 && 5 && -1 \\ -5 && 18 && -6 \\ -4 && -1 && 12 \end{pmatrix}$

Now for this case the inverse of A would be

$A^{-1} = \frac{1}{3493} \begin{pmatrix} 210 && -59 && -12 \\ 84 && 176 && 95 \\ 77 && -5 && 295 \end{pmatrix}$

Then when you multiply with the vector solution

$A^{-1} b = \frac{1}{3493} \begin{pmatrix} 210 && -59 && -12 \\ 84 && 176 && 95 \\ 77 && -5 && 295 \end{pmatrix} * \begin{pmatrix} 2400 \\ 3500 \\ 4000 \end{pmatrix} = \frac{1}{3493} \begin{pmatrix} 249500 \\ 1197600 \\ 1347300 \end{pmatrix}\\\\\\= \begin{pmatrix} \frac{500}{7} \\\\ \frac{2400}{7} \\\\ \frac{2700}{7} \end{pmatrix}$

So from that information you can conclude that the solution to the system of equations is x = 500/7 y = 2400/7 and z = 2700/7

7 0

3 years ago

6,300 and 530<br> The value of 3 _____ is ___times the value of 3 in __

inna [77]

The value of 3 in 6300 is 10 times the value of 3 in 530

7 0

3 years ago