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2 years ago

Which expression is a difference of squares with a factor of 5x − 8?

Mathematics

2 answers:

Archy [21]2 years ago

8 0

A2 - b2 = (a-b)(a+b)

(5x-8)(5x+8) = 25x2 - 64

answer: 25x2 - 64

Nitella [24]2 years ago

3 0

Answer:

option (3) is correct.

$25x^2-64$

Step-by-step explanation:

Given : factor 5x − 8

We have to choose an expression out of given option that is a difference of squares with a factor of 5x − 8

Using algebraic identity $a^2-b^2=(a+b)(a-b)$

a difference of squares is written mathematically as $a^2-b^2$

Since one factor is 5x − 8

using identity stated above other factor has be 5x+8

Thus $(5x+8)(5x-8)$

On simplify , we get

$(5x+8)(5x-8)=(5x)^2-8^2=25x^2-64$

Thus, option (3) is correct.

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The function f(x) = x2 has been translated 9 units up and 4 units to the right to form the function g(x). Which represents g(x)?

Alenkinab [10]

For this case, the parent function is given by:

$f (x) = x ^ 2$

We apply the following transformations:

Vertical translations:

Suppose that k> 0

To graph y = f (x) + k, move the graph of k units upwards:

For k = 9 we have:

$h (x) = x ^ 2 + 9$

Horizontal translations:

Suppose that h> 0

To graph y = f (x-h), move the graph of h units to the right

For h = 4 we have:

$g (x) = (x-4) ^ 2 + 9$

Answer:

The function g (x) is given by:

$g (x) = (x-4) ^ 2 + 9$

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2 years ago

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Dan divide a pie into eighths.how many equal parts are their

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There are 8 equal parts. Each part is 1/8 of the whole 8/8

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among a group of students 50 played cricket 50 played hockey and 40 played volleyball. 15 played both cricket and hockey 20 play

kondaur [170]

Answer:

Cricket only= 30

Volleyball only = 15

Hockey only = 25

Explanation:

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

From the question, we have that;

n(C) = 50, n(H) = 50, n(V) = 40

Number of students that play cricket and hockey= n(C∩H)

Number of students that play hockey and volleyball= n(H∩V)

Number of students that play cricket and volleyball = n(C∩V)

Number of students that play all three games= n(C∩H∩V)

From the question; we have,

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Therefore, number of students that play at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Thus, total number of students n(U)= 100.

Note;n(U)= the universal set

Let a = number of people who played cricket and volleyball only.

Let b = number of people who played cricket and hockey only.

Let c = number of people who played hockey and volleyball only.

Let d = number of people who played all three games.

This implies that,

d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Hence,

a = 15 – 10 = 5

b = 15 – 10 = 5

c = 20 – 10 = 10

Therefore;

For number of students that play cricket only;

n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

For number of students that play hockey only

n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

For number of students that play volleyball only

n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

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Reptile [31]

Answer:

he can expect to lose 0.5$

Step-by-step explanation:

To solve this problem we must calculate the expected value of the game.

If x is a discrete random variable that represents the gain obtained when rolling a dice, then the expected value E is:

$E =\sum xP (x)$

When throwing a dice the possible values are:

x: 1→ -$9; 2→ $4; 3→ -$9; 4→ $8; 5→ -$9; 6→ $12

The probability of obtaining any of these numbers is:

$p=\frac{1}{6}$

The gain when obtaining an even number is twice the number.

The loss to get an odd number is $ 9

So the expected gain is:

$E=-9*\frac{1}{6}-9*\frac{1}{6}-9*\frac{1}{6} + 4*\frac{1}{6} + 8*\frac{1}{6} + 12*\frac{1}{6}\\\\E =-27*\frac{1}{6} + 24*\frac{1}{6}\\\\E=-3*\frac{1}{6}\\\\E=-$0.5$

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dem82 [27]

we are given

$b+(-60)=-14$

Since, we have to solve for b

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so, we will add 60 on both sides

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so,

Answer:

Maureen should have added 60 to both sides.

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