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natta225 [31]
2 years ago
11

List all sets that -1/3 belongs to.

Mathematics
1 answer:
Vikki [24]2 years ago
5 0

Answer:

dfffffffff

Step-by-step explanation:

xcdsw23456iu76

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Devon wants to write an equation for a line that passes through 2 of the data points he has collected. The points are (8, 5) and
Svetlanka [38]
(8,5)(-12,-9)
slope = (-9 - 5) / (-12 - 8) = -14/-20 = 7/10

y = mx + b
slope(m) = 7/10
use either of ur points.....(8,5)....x = 8 and y = 5
now we sub and find b, the y int
5 = 7/10(8) + b
5 = 56/10 + b
5 - 56/10 = b
50/10 - 56/10 = b
-3/5 = b

so ur equation is : y = 7/10x - 3/5.....now we put it in standard form

y = 7/10x - 3/5...multiply by 10
10y = 7x - 6.....subtract 7x
-7x + 10y = -6....multiply by -1
7x - 10y = 6

So.....7x - 10y = 3.... IS NOT a good model. Because ur points are not a solution to ur equation.
7 0
3 years ago
Read 2 more answers
Simplify. 5√ 10 <br> 1) 2√ 2) 2√2 3) 10√2 4) 5√
Nataly_w [17]

For this case we have that the expression in its exact form is the same, that is:

5 \sqrt {10}

If it is expressed in decimal form we have:

5 \sqrt {10} = 1.5849

If we want equivalent expressions, we must first mention the following property of powers and roots:

\sqrt [n] {a ^ m} = a ^ {\frac {m} {n}}

Then, we can rewrite the expression as:

5 \sqrt {10} = \sqrt {5 ^ 2 * 10} = \sqrt {25 * 10} = \sqrt {250}

Answer:

5 \sqrt {10} = \sqrt {250} = 1.5849

8 0
3 years ago
Suppose a shoe factory produces both low-grade and high-grade shoes. The factory produces at least twice as many low-grade as hi
umka2103 [35]

Answer:

The factory should produce 166 pairs of high-grade shoes and 364 pairs of low-grade shoes for maximum profit

Step-by-step explanation:

The given parameters for the shoe production are;

The number of low grade shoes the factory produces ≥ 2 × The number of high-grade shoes produced by the factory

The maximum number of shoes the factory can produce = 500 pairs of shoes

The number of high-grade shoes the dealer calls for daily ≥ 100 pairs

The profit made per pair of high-grade shoed = Birr 2.00

The profit made per of low-grade shoes = Birr 1.00

Let 'H', represent the number of high grade shoes the factory produces and let 'L' represent he number of low-grade shoes the factory produces, we have;

L ≥ 2·H...(1)

L + H ≤ 500...(2)

H ≥ 100...(3)

Total profit, P = 2·H + L

From inequalities (1) and (2), we have;

3·H ≤ 500

H ≤ 500/3 ≈ 166

The maximum number of high-grade shoes that can be produced, H ≤ 166

Therefore, for maximum profit, the factory should produce the maximum number of high-grade shoe pairs, H = 166 pairs

The number of pairs of low grade shoes the factory should produce, L = 500 - 166 = 334 pairs

The maximum profit, P = 2 × 166 + 1 × 364 = 696

6 1
3 years ago
Read 2 more answers
Given cosx=12/13
TiliK225 [7]
1. You have the following information:

 Sinx=5/13

 Cosx=12/13

 2. By definiton, you have:

 Tanx=(Senx)/(Cosx)

 3. When you susbtitute Senx=5/13 and Cosx=12/13 into Tanx=(Senx)/(Cosx), you obtain:

 Tanx=(Senx)/(Cosx)
 Tanx=(5/13)/(12/13)
 Tanx=(5)(13)/(13)(12)
 Tanx=65/156

 4. Therefore, the answer is;

 Tanx=65/156
7 0
3 years ago
Factor 27b - 18.<br> Write your answer as a product with a whole number greater than 1.
ElenaW [278]
9(3b-2) Iv factorised it but I dont know the rest sorry
3 0
3 years ago
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