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3 years ago

If a square has a side length of x + 4 and a perimeter of 44 cm, what is the value of x. *

Mathematics

1 answer:

BARSIC [14]3 years ago

6 0

Square are the same size all the way around, which means every side = x + 4

The perimeter of a square = 4 * side

the perimeter = 44 cm

Let's plug our values into the formula above.

44 cm = 4 * (x + 4)

Distribute the 4

44 cm = 4x + 16

Subtract 16 from both sides

28 cm = 4x

Divide both sides by 4

7 = x

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For two functions, a(x) and b(x), a statement is made that a(x) = b(x) at x = 2. What is definitely true about x = 2? (2 points)

Irina-Kira [14]

The statement that is definitely true about x = 2 is Both a(x) and b(x) have the same output value at x = 2.

<h3>How to find the statement that is true about the functions a(x) = b(x) at x = 2?</h3>

If we have two functions a(x) and b(x), a statement is made that a(x) = b(x) at x = a, this implies that the values of the functions a(x) and b(x) are equal at x = a.

Given that the two functions a(x) and b(x), a statement is made that a(x) = b(x) at x = 2.

Then the statement that is definitely true about x = 2 is Both a(x) and b(x) have the same output value at x = 2.

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1 year ago

Questions (no partial grades if you don't show your work) 1. In a group of 6 boys and 4 girls, four children are to be selected.

mariarad [96]

Answer:

Total number of ways will be 209

Step-by-step explanation:

There are 6 boys and 4 girls in a group and 4 children are to be selected.

We have to find the number of ways that 4 children can be selected if at least one boy must be in the group of 4.

So the groups can be arranged as

(1 Boy + 3 girls), (2 Boy + 2 girls), (3 Boys + 1 girl), (4 boys)

Now we will find the combinations in which these arrangements can be done.

1 Boy and 3 girls = $^{6}C_{1}\times^{4}C_{3}=6\times4$ =24

2 Boy and 2 girls= $^{6}C_{2}\times^{4}C_{2}=\frac{6!}{4!\times2!}\times\frac{4!}{2!\times2!}=15\times6=90$

3 Boys and 1 girl = $^{6}C_{3}\times^{4}C_{1}=\frac{6!}{4!\times2!}\times\frac{4!}{3!}=\frac{6\times5\times4}{3 \times2} \times4=80$

4 Boys = $^{6}C_{4}=\frac{6!}{4!\times2!} =\frac{6\times 5}{2\times1}=15$

Now total number of ways = 24 + 90 + 80 + 15 = 209

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