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3 years ago

Find the solution to the system of equations.

Mathematics

1 answer:

valentinak56 [21]3 years ago

7 0

Answer:

I am not sure about this question sry but u can try asking a tutor

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Use absolute value to express the distance between −12 and −15 on the number line.

Goshia [24]

La respuesta es la por que yo ya la hize D

4 0

3 years ago

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Brain if correct help pls

zheka24 [161]

Step-by-step explanation:

<u>Complete the operation, find the quotient and match with correct option:</u>

44.1/4.9 = 9
32.8/8.2 = 4
43.2/3.6 = 12
161.2/5.2 = 31
93.6/7.2 = 13

3 0

2 years ago

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The Falcons had 3 fewer than twice as many Bears players. If the Falcons had 11 players, how many players were on the Bears?

Dmitry [639]

Subtraction, 11-3 is 8
Ps:trick question

3 0

2 years ago

Use the graph to find the following.

Levart [38]

a) Local maximum values of g are: 0 and 3

b) The values at which g has local maximum are : -2 and 2

<h3>What is a maximum point?</h3>

A maximum point is the point at which the gradient of the curve changes from positive to negative.

Analysis:

The inverted v-curves all have maximum values and they are two.

The one at x = -2 and the one at x = 2 for in both cases their maximum values are y = 0 and y = 3 respectively.

Learn more about minimum and maximum points: brainly.com/question/14993153

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2 years ago

1)How many ways can the letters in the word BOOKKEEPER be arranged? (Must show set-up )

sergeinik [125]

Question 1

There are 5 letters (B, O, K, E, R) and there is a total of 10 letters to make up the word.
There are $\frac{10!}{(10-6)!6!}$ ways of arranging the letters, which equal to 210 ways

Question 2

There are seven swimmers in total.
There are $\frac{7!}{(7-1)!1!}$ ways of choosing the first winner, which is 7 ways
There are $\frac{6!}{(6-1)!1!}$ ways of choosing the second winner, which is 6 ways
There are $\frac{5!}{(5-1)!1!}$ ways of choosing the third winner, which is 5 ways
There are 7×6×5=210 ways of choosing first, second, and third winner

Question 3

The probability of eating an orange and a red candy is $\frac{15}{31}$ × $\frac{9}{30}$ , which equals to $\frac{9}{62}$

The probability of eating two green candies is $\frac{7}{31}$ × $\frac{6}{30}$ which equals to $\frac{7}{155}$

3 0

3 years ago