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3 years ago

Re 6. Sketch a net that will not result in a cube.

Mathematics

1 answer:

sergey [27]3 years ago

4 0

Answer:

One line then the next then repeat

hahahahah

Step-by-step explanation:

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OMGGGGGGG PLSSS HELP ME WITH THIS QUESTION, IM SUPER STUCK PLS PLS PLS ITS DUE VERY SOON!!!! EXPLAIN PLSSS, WILL MARK THE BEST A

siniylev [52]

It’s H. You have to set up an equation: y=10x-11. -11 is the starting value, so you put it at the end. You add ten each time, so you put 10x. Just plug in 100 for x, and you’ll get 989

4 0

3 years ago

In the expansion of (1/ax +2ax^2)^5 the coefficient of x is five. Find the value of the constant a.

DedPeter [7]

Answer:

80x⁴

Step-by-step explanation:

$(\frac{1}{ax} + 2ax^2)^5 = 5C_0(\frac{1}{ax})^5(2ax^2)^0 + 5C_1(\frac{1}{ax})^4(2ax^2)^1 + 5C_2(\frac{1}{ax})^3 (2ax^2)^2$

$+ 5C_3 (\frac{1}{ax})^2(2ax^2)^3 + 5C_4(\frac{1}{ax})^1(2ax^2)^4 + 5C_5(\frac{1}{ax})^0(2ax^2)^5$

$5C_0(\frac{1}{ax})^5(2ax^2)^0 =1 \times (\frac{1}{ax})^5 \times 1 = \frac{1}{a^5x^5}\\\\5C_1(\frac{1}{ax})^4(2ax^2)^1 = 5 \times (\frac{1}{ax})^4 \times (2ax^2)^1 = 10 ax^2 \times \frac{1}{a^4x^4} = \frac{10}{a^3x^2}\\\\5C_2 (\frac{1}{ax})^3 (2ax^2)^2= 10 \times (\frac{1}{ax})^3 \times (2ax^2)^2 = 10 \times \frac{1}{a^3x^3} \times 4a^2x^4 = \frac{40x}{a}\\\\5C_3 (\frac{1}{ax})^2 (2ax^2)^3 = 10 \times (\frac{1}{ax})^2 \times (2ax^2)^3 = 10 \times \frac{1}{a^2x^2} \times 8a^3 x^6 = 80ax^4\\\\$

$5C_4(\frac{1}{ax})^1(2ax^2)^4 = 5 \times \frac{1}{ax} \times 16a^4x^8 = 80a^3x^7\\\\5C_5(\frac{1}{ax})^0(2ax^2)^5 = 1 \times 1 \times 32a^5x^{10}$

The fourth term of the expansion has the constant a,

the coefficient of a is 80x⁴

6 0

3 years ago

Round 24,732 to the<br> nearest thousand.

kherson [118]

Answer:

25,000

Step-by-step explanation:

Since 7 is greater than 5 the 4 rounds up to a 5

4 0

4 years ago

Read 2 more answers

Of 75is90 =120 true or false

Mariana [72]

Answer:

False

Step-by-step explanation:

8 0

3 years ago

In a probability experiment, Craig rolled a six-sided die 57 times. The die landed on a number greater than three 33 times. What

GaryK [48]

The ratio is 33:24 which would simplify to 11:8.

There were 33 rolls that were greater than 33. Since there were 57 rolls altogether, there were 57-33 = 24 rolls that were not greater than 33; they must be less than or equal to 33. Thus the ratio is 33:24.

6 0

4 years ago