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steposvetlana [31]
3 years ago
6

Write 40/64 in simplest

Mathematics
1 answer:
Leviafan [203]3 years ago
5 0
5/8 is the simplified fraction for 40/64
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What is the 82nd term of the sequence below?
AlexFokin [52]
The rule to find the term is: y=225-4x
you want to know the 82nd term which occurs when x=82 therefore
y = 225-(4 x 82) = -103


82nd term = -103
4 0
3 years ago
Convert a fraction to a decimal number . <br> 8 divided by 3.000. PLEASE HELP ME !!!!!!
RoseWind [281]

Answer:

2.67

Step-by-step explanation:

7 0
3 years ago
Find the surface area of the right square pyramid. Round your answer to the nearest hundredth.
jekas [21]

Answer:

A. 176

Step-by-step explanation: First, to find the surface area to each triangle, you multiply the base times the height by 1/2. It is 28 for each triangle. Multiply that by 4, and you get 112. Then, the surface area of the square on bottom is 64. When added together, you get 176. No rounding needed. Hope it helped and is correct!

7 0
3 years ago
Someone help please!
trapecia [35]

Answer:

x = 6

Step-by-step explanation:

The sum of the angles of a triangle is 180

70+60+8x+2 = 180

Combine like terms

132 +8x = 180

Subtract 132 from each side

132+8x-132= 180-132

8x = 48

Divide each side by 8

8x/8 = 48/8

x = 6

6 0
3 years ago
The ideal width of a safety belt strap for a certain automobile is 6 cm. The actual width can vary by at most 0.45 cm. Write an
OlgaM077 [116]

Answer:

|x-6|\le 0.45

x\in [5.55,6.45]

Step-by-step explanation:

<u>Absolute Value Inequality</u>

Assume the actual width of a safety belt strap for a certain automobile is x. We know the ideal width of the strap is 6 cm. This means the variation from the ideal width is x-6.

Note if x is less than 6, then the variation is negative. We usually don't care about the sign of the variation, just the number. That is why we need to use the absolute value function.

The variation (unsigned) from the ideal width is:

|x-6|

The question requires that the variation is at most 0.45 cm. That poses the inequality:

|x-6|\le 0.45

That is the range of acceptable widths. Let's now solve the inequality.

To solve an inequality for an absolute value less than a positive number N, we write:

-0.45\le x-6 \le 0.45

This is a double inequality than can be easily solved by adding 6 to all the sides.

-0.45+6\le x \le 0.45+6

Operating:

5.55\le x \le 6.45

That is the solution in inequality form. Expressing in interval form:

\boxed{x\in [5.55,6.45]}

3 0
4 years ago
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