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Ket [755]
3 years ago
11

If l x w = v, what is the value of v?

Mathematics
1 answer:
stepladder [879]3 years ago
3 0
The value of v would be the values of the length times times the width.
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Pls help me complete this
snow_tiger [21]

Answer: The number of times a yellow cube will be drawn is 390 times.

Step-by-step explanation: Like the last question with the blue cubes, set up a ratio.

Since it is still 240 times drawn and we need to figure out how many times a yellow cube will be drawn from 1200 times, the top number only changes.

78/240 = x/1200

Multiply 78 and 1200, which is 93,600.

Divide that by 93,600, which is 390.

7 0
3 years ago
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dimaraw [331]
B I think please give me brainliest
6 0
3 years ago
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Simplify. (x6 y3)(−8xy2)(−6x3)
meriva
When simplifying this, we would have to take this little by little in serious steps and not only this, but to also take notice of the use of <u>pemdas</u> and to make sure that we take each step carefully.

We first would <u></u><u>(first)</u> simplify the following:

(((x^6)*(y^3))*(0-(8x*(y^2))))*(0-(2*3x^3)) \\ \\ then \ we \ would \ do . .  . . \\ \\  x^7  \ multiplied \ by \ x^3 \ = x^(^7^ +^ 3^) = \ x^1^0 \\ \\ (last \ step).... \\ \\ 2^3 \   multiplied \ by \ 2^1   = 2^(^3^ +^ 1^) = 2^4 \\ \\ \Longrightarrow \ (your \ answer) \\ \\ \boxed{\bf{  (2^4*3x^1^0y^5)}}
7 0
3 years ago
What expression simplifies to 2i ? help asap lots of points
PilotLPTM [1.2K]

Answer:

Step-by-step explanation:

First, consider the following expression.

(6x + 8) + (4x + 2)

To simplify this expression, you combine the like terms, 6x and 4x. These are like terms because they have the same variable with the same exponents. Similarly, 8 and 2 are like terms because they are both constants, with no variables.

(6x + 8) + (4x + 2) = 10x + 10

5 0
2 years ago
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Simplify the polynomial expressions by combining like terms and then multiply the resulting binomial expressions to find their p
patriot [66]

(6x-9)(8+5x-5)=(6x-9)(5x+3)=

=6x\cdot5x-5x\cdot9+6x\cdot3-9\cdot3=\\=30x^2-27x-27


6 0
3 years ago
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