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Gelneren [198K]
3 years ago
15

– 3х – бу = 17; (6, 3)

Mathematics
1 answer:
almond37 [142]3 years ago
8 0

I'm going to assume that you want to know whether the coordinate satisfies the equation. Let's find out;

-3х - 6у = 17

-3(6) - 6(3) = 17

-18 - 18 = 17

-36 ≠ 17

The point (6, 3) does NOT satisfy the equation.

Best of Luck!

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Solve the inequality. Show your work.<br> |4r + 8| ≥ 32
saw5 [17]

|4r + 8| ≥ 32

4r + 8 ≥ 32         or         4r + 8 ≤ -32      <em> inside of absolute value could be + or -</em>

4r       ≥ 24         or          4r       ≤ -40       <em>subtracted 8 from both sides</em>  

r       ≥  6           or            r       ≤ -10         <em>divided both sides by 4</em>

Graph: ←---------------- -10        6 ------------------→

Interval Notation: (-∞, -10] U [6, ∞)

7 0
3 years ago
Read 3 more answers
Rami buys 3 pounds of oranges . If the cost is represented by c, what is the cost per pound can you define the variable
SCORPION-xisa [38]
I believe it’s 3 over c.
3 0
2 years ago
f(x)= 1 / x-3 g(x)= 3x + 1/x A. Use composition to prove whether or not the functions are inverses of each other. B. Express the
Alisiya [41]

Step-by-step explanation:

(A) To find inverse of f(x) = y = 1 / x - 3 :

<u>Transform in terms of x</u>

y = 1 / x - 3

x - 3 = 1 / y

x = 3 + 1 / y

<u>Exchange x with y</u> :

{f}^{ - 1} (x) = 3 +  \frac{1}{x}

Therefore, g(x) is not inverse of f(x) and vice versa.

(B)

Domain of f(x) = R - {3}

Domain of g(x) = R - {0}

7 0
3 years ago
1.How many combinations are possible from the letters HOLIDAY if the letters are taken?
Lesechka [4]

Using the combination formula, it is found that:

1. A. 7 combinations are possible.

B. 21 combinations are possible.

C. 1 combination is possible.

2. There are 245 ways to group them.

<h3>What is the combination formula?</h3>

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by:

C_{n,x} = \frac{n!}{x!(n-x)!}

Exercise 1, item a:

One letter from a set of 7, hence:

C_{7,1} = \frac{7!}{1!6!} = 7

7 combinations are possible.

Item b:

Two letters from a set of 7, hence:

C_{7,2} = \frac{7!}{2!5!} = 21

21 combinations are possible.

Item c:

7 letters from a set of 7, hence:

C_{7,7} = \frac{7!}{0!7!} = 1

1 combination is possible.

Question 2:

Three singers are taken from a set of 7, and four dances from a set of 10, hence:

T = C_{7,3}C_{10,4} = \frac{7!}{3!4!} \times \frac{10!}{4!6!} = 245

There are 245 ways to group them.

More can be learned about the combination formula at brainly.com/question/25821700

6 0
2 years ago
Find the measure of c in each figure below using the side lengths given
kiruha [24]

Answer:

Step-by-step explanation:

take angle C as reference angle

using cos rule

cos C=adjacent/hypotenuse

cos C=21/75

cos C=0.28

C=cos^{-1}0.28

C=73.7 degree

3 0
3 years ago
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