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snow_lady [41]
3 years ago
12

The sum of three consecutive odd numbers is 159. What is the smallest of the three numbers ?

Mathematics
1 answer:
natta225 [31]3 years ago
6 0

Since we don't know where to start, we can always give that a variable which in this case, it's X. Since we are finding the sum of three consecutive odd numbers, we will be skipping numbers so we can add 2 each time. Now, let's set up an equation.

<h2>X + X + 2 + X + 4 = 159</h2><h2 /><h2>3x + 6 = 159</h2><h2>       -6     -6   ← subtract 6 on both sides</h2><h2 /><h2>       3x = 153</h2><h2>       ÷3     ÷3   ← divide both sides by 3</h2><h2 /><h2>          X = 51</h2><h2>          X + 2 = 53</h2><h2>          X + 4 = 55</h2><h2 />

Therefore, the smallest of our three consecutive integers is 51.

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Max's Market sells 6 eggs for £2.04.
nlexa [21]

Answer:

Groceries Galore

Step-by-step explanation:

1) Find the value of 1 egg for each shop.

<u>Max's Market: </u>

2.04 ÷ 6

= £0.34

<u>Groceries Galore:</u>

2.64 ÷ 8

= £0.33

<u>Fancy Foodstuffs:</u>

4.08 ÷ 12

= £0.34

Therefore, Groceries Galore's price per egg is cheaper.

4 0
2 years ago
Read 2 more answers
Bananas are on sale for $0.39 per pound. Mr Schurter bought 3 x 3 /4 pounds of bananas. Which is closest to the amount he paid f
Anna71 [15]
It comes out to 0.8775 so it rounds to about $0.88
8 0
3 years ago
Help me with solution please =(​
Slav-nsk [51]

\huge \boxed{\mathfrak{Question} \downarrow}

Factorise the polynomials.

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

<h3><u>1. b² + 8b + 7</u></h3>

\sf \: b ^ { 2 } + 8 b + 7

Factor the expression by grouping. First, the expression needs to be rewritten as b²+pb+qb+7. To find p and q, set up a system to be solved.

\sf \: p+q=8 \\  \sf pq=1\times 7=7

As pq is positive, p and q have the same sign. As p+q is positive, p and q are both positive. The only such pair is the system solution.

\sf \: p=1  \\  \sf \: q=7

Rewrite \sf\:b^{2}+8b+7 \: as \: \left(b^{2}+b\right)+\left(7b+7\right).

\sf \: \left(b^{2}+b\right)+\left(7b+7\right)

Take out the common factors.

\sf \: b\left(b+1\right)+7\left(b+1\right)

Factor out common term b+1 by using distributive property.

\boxed{\boxed{ \bf \: \left(b+1\right)\left(b+7\right) }}

__________________

<h3><u>2. 4x² + 4x + 1</u></h3>

\sf \: 4 x ^ { 2 } + 4 x + 1

Factor the expression by grouping. First, the expression needs to be rewritten as 4x²+ax+bx+1. To find a and b, set up a system to be solved.

\sf \: a+b=4 \\  \sf ab=4\times 1=4

As ab is positive, a and b have the same sign. As a+b is positive, a and b are both positive. List all such integer pairs that give product 4.

\sf \: 1,4 \\  \sf 2,2

Calculate the sum for each pair.

\sf \: 1+4=5  \\  \sf \: 2+2=4

The solution is the pair that gives sum 4.

\sf \: a=2 \\  \sf b=2

Rewrite \sf4x^{2}+4x+1 as \left(4x^{2}+2x\right)+\left(2x+1\right).

\sf \: \left(4x^{2}+2x\right)+\left(2x+1\right)

Factor out 2x in 4x² + 2x.

\sf \: 2x\left(2x+1\right)+2x+1

Factor out common term 2x+1 by using distributive property.

\sf\left(2x+1\right)\left(2x+1\right)

Rewrite as a binomial square.

\boxed{ \boxed{\bf\left(2x+1\right)^{2}} }

__________________

<h3><u>3. 5n² + 10n + 20</u></h3>

\sf \: 5 n ^ { 2 } + 10 n + 20

Factor out 5. Polynomial n² + 2n+4 is not factored as it does not have any rational roots.

\boxed{\boxed{\bf5\left(n^{2}+2n+4\right)} }

<u>___________</u>_______

<h3><u>4. m³ - 729</u></h3>

\sf \: m ^ { 3 } - 729

Rewrite m³-729 as m³ - 9³. The difference of cubes can be factored using the rule: a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right). Polynomial m²+9m+81 is not factored as it does not have any rational roots.

\boxed{ \boxed{ \bf \: \left(m-9\right)\left(m^{2}+9m+81\right) }}

__________________

<h3><u>5. x² - 81</u></h3>

\sf \: x ^ { 2 } - 81

Rewrite x²-81 as x² - 9². The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

\boxed{ \boxed{ \bf\left(x-9\right)\left(x+9\right) }}

__________________

<h3><u>6. 15x² - 17x - 4</u></h3>

\sf \: 15 x ^ { 2 } - 17 x - 4

Factor the expression by grouping. First, the expression needs to be rewritten as 15x²+ax+bx-4. To find a and b, set up a system to be solved.

\sf \: a+b=-17 \\   \sf \: ab=15\left(-4\right)=-60

As ab is negative, a and b have the opposite signs. As a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -60.

\sf \: 1,-60  \\ \sf 2,-30  \\  \sf3,-20  \\  \sf4,-15 \\   \sf5,-12  \\ \sf 6,-10

Calculate the sum for each pair.

\sf \: 1-60=-59  \\ \sf 2-30=-28 \\   \sf3-20=-17 \\   \sf4-15=-11  \\  \sf5-12=-7  \\  \sf6-10=-4

The solution is the pair that gives sum -17.

\sf \: a=-20  \\  \sf \: b=3

Rewrite \sf15x^{2}-17x-4 as \sf \left(15x^{2}-20x\right)+\left(3x-4\right).

\sf\left(15x^{2}-20x\right)+\left(3x-4\right)

Factor out 5x in 15x²-20x.

\sf \: 5x\left(3x-4\right)+3x-4

Factor out common term 3x-4 by using distributive property.

\boxed{\boxed{ \bf\left(3x-4\right)\left(5x+1\right) }}

6 0
3 years ago
Read 2 more answers
Solve: 3 x (8 + 2) ÷ 2
kobusy [5.1K]

Answer:

15

Step-by-step explanation:

3 ×(8 +2) :2 =

3 ×10 :2 =

30 :2 =

15

8 0
3 years ago
1) Ms. Kitts works at a music store. Last week she sold 6 more than 3 times the number of CDs that she sold this week. Ms. Kitts
Alborosie

Answer:

L = 3t + 6

108 = L + t

Step-by-step explanation:

first, define the variables:

L = # of CDs sold last week

t = # of CDs sold this week

then, let's write what the words of problem tells us in numbers:

"Last week (L), she sold 6 more (+ 6) than 3 times the number of CDs she sold this week (3t)

L = 3t + 6

"She sold a total of 108 CDs over 2 weeks (L + t)

108 = L + t

7 0
3 years ago
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