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chubhunter [2.5K]
3 years ago
10

A builder needs to cover 10 square yards of floor with tile. red tile costs $60 per square yard, and white tole costs $40 per sq

uare yard. if the builder spends exactly $480 on tile, how much of the floor will be overed with RED tile?
Mathematics
1 answer:
dsp733 years ago
6 0

Answer:40 square yards

Step-by-step explanation:

Given

Floor area A=10 square yards

Red tile costs \$ 60 per square yards

White tile costs \$40 per square yards

Total money spent=\$480 on tiles

Suppose red tiles cover an area of x square yards

So, we can write

60\times x+40\times (10-x)=480\\60x+400-40x=480\\20x=80\\x=40\ \text{square yards}

Red tiles cover an area of 40 square yards  

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Look ate the picture down below, and please help me i would really appreciate it
zavuch27 [327]

Answer:

  46°

Step-by-step explanation:

When secants intersect each other and a circle, the external angle (A) is half the difference of the intercepted arcs:

  ∠A = (arcDC -arcBC)/2

  12° = (arcDC -22°)/2 . . . . . . . fill in the given numbers

  24° = arcDC -22° . . . . . . . . . multiply by 2

  46° = arcDC . . . . . . . . . . . . . add 22°

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%24a%2Ba%20r%2Ba%20r%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D15%24%24a%5E%7B2%7D%2B%28a%20r%29%5E%7B
riadik2000 [5.3K]

Let

S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n

where we assume |r| < 1. Multiplying on both sides by r gives

r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}

and subtracting this from S_n gives

(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}

As n → ∞, the exponential term will converge to 0, and the partial sums S_n will converge to

\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}

Now, we're given

a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a

a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}

We must have |r| < 1 since both sums converge, so

\dfrac{15}a = \dfrac1{1-r}

\dfrac{150}{a^2} = \dfrac1{1-r^2}

Solving for r by substitution, we have

\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)

\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}

Recalling the difference of squares identity, we have

\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}

We've already confirmed r ≠ 1, so we can simplify this to

\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15

It follows that

\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12

and so the sum we want is

ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

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2 years ago
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Fed [463]

Answer:

2+10x-x^2, the last answer

Step-by-step explanation:

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3 years ago
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Answer:

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3 years ago
The left and right pages of an open book are two consecutive integers whose sum is 359. Find these page numbers
ZanzabumX [31]

179 and 180

consecutive numbers have a difference of 1 between them

let the 2 numbers be n and n + 1, then

n + n + 1 = 359

2n + 1 = 359 ( subtract 1 from both sides )

2n = 358 ( divide both sides by 2 )

n = 179 and n + 1 = 179 + 1 = 180

the 2 pages are 179 and 180



6 0
3 years ago
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