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Pie
2 years ago
12

Mr. Pete just got a great deal at Lowe’s on tiles for his new patio. He wants to

Mathematics
1 answer:
sdas [7]2 years ago
8 0

Answer:

The fewest square tiles he can use without cutting any of them = 12 square tiles

Step-by-step explanation:

The given dimensions of Mr. Pete's rectangular patio = 108 inches by 144 inches

The area of the patio = 108 in. × 144 in. = 15,552 in.²

We note that when square tiles are used the must fit both the length and the with of the rectangular patio

Therefore, we have;

The dimensions of the square tiles that the fewest number of tiles is the highest common factor, HCF, of 108 inches and 144

The HCF of 108 and 144 are found from by the prime factors as follows;

108 = 2² × 3³ = 2² × 3² × 3

144 = 2⁴ × 3² = 2² × 3² × 4

From which we have;

The HCF of 108 and 144 = 2² × 3² = 36

Therefore, the size of the sides of the square tiles that will give the fewest number of tiles is 36 inches

The number of the 36 inches square tiles that fit into the length of the rectangular patio = 144 inches/(36 inches/tile) = 4 tiles

The number of the 36 inches square tiles that fit into the breadth of the rectangular patio = 108 inches/(36 inches/tile) = 3 tiles

The total number of tiles he will use = (4 × 3) tiles = 12 tiles = The fewest square tiles he can use without cutting any of them.

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castortr0y [4]

Answer:

216 cubic inches

Step-by-step explanation:

(9)(3)(3)=81

(3)(3)(9)=81

(3)(3)(6)=54

81+81+54=216

So, it's 216 cubic inches in volume.

6 0
3 years ago
If (ax+2)(bx+7)=15x2+cx+14 for all values of x, and a+b=8, what are the 2 possible values fo c
dolphi86 [110]

Given:

(ax+2)(bx+7)=15x^2+cx+14

And

a+b=8

Required:

To find the two possible values of c.

Explanation:

Consider

\begin{gathered} (ax+2)(bx+7)=15x^2+cx+14 \\ abx^2+7ax+2bx+14=15x^2+cx+14 \end{gathered}

So

\begin{gathered} ab=15-----(1) \\ 7a+2b=c \end{gathered}

And also given

a+b=8---(2)

Now from (1) and (2), we get

\begin{gathered} a+\frac{15}{a}=8 \\  \\ a^2+15=8a \\  \\ a^2-8a+15=0 \end{gathered}a=3,5

Now put a in (1) we get

\begin{gathered} (3)b=15 \\ b=\frac{15}{3} \\ b=5 \\ OR \\ b=\frac{15}{5} \\ b=3 \end{gathered}

We can interpret that either of a or b are equal to 3 or 5.

When a=3 and b=5, we have

\begin{gathered} c=7(3)+2(5) \\ =21+10 \\ =31 \end{gathered}

When a=5 and b=3, we have

\begin{gathered} c=7(5)+2(3) \\ =35+6 \\ =41 \end{gathered}

Final Answer:

The option D is correct.

31 and 41

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