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Volgvan
3 years ago
5

(5a - 5b + 7) - (2a - 5b - 3)

Mathematics
2 answers:
never [62]3 years ago
6 0

This equals 3a+10

Hope this helps!

Fofino [41]3 years ago
4 0
<span>(5a - 5b + 7) - (2a - 5b - 3) = 5a - 5b + 7 - 2a + 5b + 3 = 3a + 10 

</span>
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What is the product of -9(5-2x)
nata0808 [166]
-45+18x is the answer
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5 0
4 years ago
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Thanks in advance for your answer :)
MAXImum [283]

Answer:

x=5

Step-by-step explanation:

its easy just subtract 9 from 24 to get 15 and 15 divided by 3 is 5

7 0
4 years ago
A population of bacteria is 8500 on Day 1, 9350 on Day 2, and 10285 on Day 3.
11111nata11111 [884]

Answer:

<em>The population on 30th day will be </em><em>134836</em><em>.</em>

Step-by-step explanation:

This is a case of exponential growth. The general form of exponential function is,

y=ab^x

where, a and b are constants.

The data points from the question are (1,8500),(2,9350),(3,10285)

Putting the values in the function,

for x=1, y=8500

8500=ab  ----------1

for x=2, y=9350

9350=ab^2  ------2

Dividing equation 2 by 1,

\Rightarrow \dfrac{ab^2}{ab}=\dfrac{9350}{8500}=1.1

\Rightarrow b=1.1

Putting the value of b in equation 1,

\Rightarrow a=\dfrac{8500}{1.1}=7727.27

Now the function becomes,

y=7727.27(1.1)^x

Putting x=30, we get

y=7727.27(1.1)^{30}=134836.2\approx 134836

4 0
3 years ago
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Find all numbers whose absolute value is 4
Elan Coil [88]
The answer would be -4 and 4
8 0
3 years ago
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You flip a coin and then roll a 6-sided number cube (a die).
expeople1 [14]

Answer:

a)  No, it does not matter whether you roll the die or flip the coin first, as these two events are <u>independent</u> of each other, which means they do not affect each other.

b) Yes.

  • Let event 1 be flipping a coin and event 2 be rolling a die.
  • Let event 1 be rolling a die and event 2 be flipping a coin.

The likelihood that any outcome will occur will not change, as the events are independent.

c) see attached

d)   12 outcomes  (H = head, T = tail, numbers represent the value of the die)

H 1           T 1

H 2          T 2

H 3          T 3

H 4          T 4

H 5          T 5

H 6          T 6

e)  

\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}

\implies \sf P(even)=\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{3}{6}=\dfrac{1}{2}

\implies \sf P(head)=\dfrac{1}{2}

\implies \sf P(even)\:and\:P(head)=\dfrac{1}{2} \times \dfrac{1}{2}=\dfrac{1}{4}

6 0
2 years ago
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