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sp2606 [1]
3 years ago
7

What is the midpoint between 0.13 and 0.14

Mathematics
1 answer:
weqwewe [10]3 years ago
3 0
0.13 + 0.14= 0.27 0.27 ÷ 2 = 0.135 Therefore the midpt. Is 0.135
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The price of an item has been reduced by 10%. The original price was $51. What is the prince of the item now?
AlekseyPX

Answer:

The new price is 45.90

Step-by-step explanation:

The original price is 51

Find the discount of 10 %

51*10%

51*.10

5.1

Subtract this from the original price

51- 5.1

45.9

The new price is 45.90

4 0
2 years ago
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H(x) = 3x-4 what is h(6)
lina2011 [118]

Answer:

14

Step-by-step explanation:

3*6=18

18-4=14

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Given that 3x+2y=5a+b and 4x-3y=a+7b, solve for y-x in terms of a and b.​
Eduardwww [97]

Answer:

y - x = - 2 b

Step-by-step explanation:

Given-

3 x + 2 y = 5 a + b...................(i)

4 x - 3 y = a + 7 b.....................(ii)

<em><u>Multiply by 3 and 2 in equation (i) & (ii) respectively-</u></em>

<em><u>Add equation (iii) &  (iv)-</u></em>

9 x + 6 y = 15 a + 3 a...................(iii)

<u>8 x - 6 y = 2 a + 14 b</u> ...................(iv)

17 x = 17 a + 17 b

<u>x = a + b</u>

Put x = a + b in equation (i)-

<em>2 y = 5 a + b - 3 a - 3 b</em>

<em>2 y = 2 a - 2 b</em>

<u>y = a - b </u>

<em>∴ y - x = a - b - ( a + b )</em>

<em>y - x = a - b - a - b</em>

y - x = - 2 b

3 0
3 years ago
Which expressions are equivalent to 2(2x + 4y + x − 2y)?
Greeley [361]
C should be your answer:))))
5 0
3 years ago
Sphere A is similar to Sphere B. The scale factor of the lengths of the radii of Sphere A to Sphere B is 1 to 4. Sphere A has th
AleksAgata [21]

\bf ~\hspace{5em} \textit{ratio relations of two similar shapes} \\\\ \begin{array}{ccccllll} &\stackrel{\stackrel{ratio}{of~the}}{Sides}&\stackrel{\stackrel{ratio}{of~the}}{Areas}&\stackrel{\stackrel{ratio}{of~the}}{Volumes}\\ \cline{2-4}&\\ \cfrac{\stackrel{similar}{shape}}{\stackrel{similar}{shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}~\hspace{6em} \cfrac{s}{s}=\cfrac{\sqrt{Area}}{\sqrt{Area}}=\cfrac{\sqrt[3]{Volume}}{\sqrt[3]{Volume}} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \cfrac{\textit{sphere A}}{\textit{sphere B}}\qquad \stackrel{\stackrel{sides'}{ratio}}{\cfrac{1}{4}}\qquad \qquad \stackrel{\stackrel{sides'}{ratio}}{\cfrac{1}{4}}=\stackrel{\stackrel{volumes'}{ratio}}{\cfrac{\sqrt[3]{288}}{\sqrt[3]{v}}}\implies \cfrac{1}{4}=\sqrt[3]{\cfrac{288}{v}}\implies \left( \cfrac{1}{4} \right)^3=\cfrac{288}{v} \\\\\\ \cfrac{1^3}{4^3}=\cfrac{288}{v}\implies \cfrac{1}{64}=\cfrac{288}{v}\implies v=18432

3 0
3 years ago
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