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3 years ago

What is the factored form of 9x^2-16

Mathematics

1 answer:

Scilla [17]3 years ago

7 0

Answer:

Hey! The answer to you're question is (3x+4) (3x+4)

Step-by-step explanation:

The best way of how to solve this problem is break it up.

If you know how to square root, that would come in handy!

The square root of 9 is 3! The square root of 16 is 4.

The ^2 is just telling you that there are two variables and you would need to split them. :0

(a square root is 2 of the same numbers that equal something, like 6x6=36 ect.)

so 9x^2-16

3x+4

Therefore!!!!!! (3x+4) x (3x+4)

Dont feel bad for not asking :) we all get stuck sometimes, have a good rest of you're night sweetie!

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Radda [10]

Answer:

1. D

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4. A

5. A

6. F

Step-by-step explanation:

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3 years ago

HELP…this is probably easy…but still help.

andreev551 [17]

Answer:

A) The parallelogram is a rhombus

Step-by-step explanation:

A rhombus is an equally 4-sided figure with two acute angles and two obtuse angles opposite of each other.

4 0

2 years ago

A total of 12 players consisting 6 male and 6 female badminton players are attending a training camp

abruzzese [7]

Step-by-step explanation:

"A total of 12 players consisting 6 male and 6 female badminton players are attending a training camp."

"(a) During a morning activity of the camp, these 12 players have to randomly group into six pairs of two players each."

"(i) Find the total number of possible ways that these six pairs can be formed."

The order doesn't matter (AB is the same as BA), so use combinations.

For the first pair, there are ₁₂C₂ ways to choose 2 people from 12.

For the second pair, there are ₁₀C₂ ways to choose 2 people from 10.

So on and so forth. The total number of combinations is:

₁₂C₂ × ₁₀C₂ × ₈C₂ × ₆C₂ × ₄C₂ × ₂C₂

= 66 × 45 × 28 × 15 × 6 × 1

= 7,484,400

"(ii) Find the probability that each pair contains players of the same gender only. Correct your final answer to 4 decimal places."

We need to find the number of ways that 6 boys can be grouped into 3 pairs. Using the same logic as before:

₆C₂ × ₄C₂ × ₂C₂

= 15 × 6 × 1

= 90

There are 90 ways that 6 boys can be grouped into 3 pairs, which means there's also 90 ways that 6 girls can be grouped into 3 pairs. So the probability is:

90 × 90 / 7,484,400

= 1 / 924

≈ 0.0011

"(b) During an afternoon activity of the camp, 6 players are randomly selected and 6 one-on-one matches with the coach are to be scheduled.

(i) How many different schedules are possible?"

There are ₁₂C₆ ways that 6 players can be selected from 12. From there, each possible schedule has a different order of players, so we need to use permutations.

There are 6 options for the first match. After that, there are 5 options for the second match. Then 4 options for the third match. So on and so forth. So the number of permutations is 6!.

The total number of possible schedules is:

₁₂C₆ × 6!

= 924 × 720

= 665,280

"(ii) Find the probability that the number of selected male players is higher than that of female players given that at most 4 females were selected. Correct your final answer to 4 decimal places."

If at most 4 girls are selected, that means there's either 0, 1, 2, 3, or 4 girls.

If 0 girls are selected, the number of combinations is:

₆C₆ × ₆C₀ = 1 × 1 = 1

If 1 girl is selected, the number of combinations is:

₆C₅ × ₆C₁ = 6 × 6 = 36

If 2 girls are selected, the number of combinations is:

₆C₄ × ₆C₂ = 15 × 15 = 225

If 3 girls are selected, the number of combinations is:

₆C₃ × ₆C₃ = 20 × 20 = 400

If 4 girls are selected, the number of combinations is:

₆C₂ × ₆C₄ = 15 × 15 = 225

The probability that there are more boys than girls is:

(1 + 36 + 225) / (1 + 36 + 225 + 400 + 225)

= 262 / 887

≈ 0.2954

7 0

3 years ago

\ Which center of a triangle is also the center of gravity of the triangle? orthocenter centroid incenter circumcenter

Citrus2011 [14]

That is the centroid.

The point where the lines from each vertex of the triangle to the midpoints of the opposite sides intersect.

3 0

3 years ago

Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into

Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = $(\frac{60\pi }{\pi+4}) in$

b)the length of the wire for the square = $(\frac{240}{\pi+4}) in$

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

b= $\frac{y}{4}$

Area of the square which is L² can now be said to be;

$A_S=(\frac{y}{4})^2 = \frac{y^2}{16}$

On the otherhand; let the radius (r) of the circle be;

2πr = 60-y

$r = \frac{60-y}{2\pi }$

Area of the circle which is πr² can now be;

$A_C= \pi (\frac{60-y}{2\pi } )^2$

$=( \frac{60-y}{4\pi } )^2$

Total Area (A);

A = $A_S+A_C$

= $\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2$

For the smallest possible area; $\frac{dA}{dy}=0$

∴ $\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0$

If we divide through with (2) and each entity move to the opposite side; we have:

$\frac{y}{18}=\frac{(60-y)}{2\pi}$

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

$y= \frac{240}{\pi+4}$

∴ since $y= \frac{240}{\pi+4}$ , we can determine for the length of the circle ;

60-y can now be;

= $60-\frac{240}{\pi+4}$

= $\frac{(\pi+4)*60-240}{\pi+40}$

= $\frac{60\pi+240-240}{\pi+4}$

= $(\frac{60\pi}{\pi+4})in$

also, the length of wire for the square (y) ; $y= (\frac{240}{\pi+4})in$

The smallest possible area (A) = $\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})$

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

4 0

4 years ago