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Elis [28]
3 years ago
13

7 of 9

Mathematics
1 answer:
lidiya [134]3 years ago
5 0
Answer- 2\15 as it is by the white fabric
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10 + (2 x 3)2 ÷ 4 × 1 over 2 to the power of 3
diamong [38]

Answer:

The admixture of words and math symbols must be rewritten to allow processing towards a correct answer.

Step-by-step explanation:

Use brackets [ ] and up carrot symbol (^, obtained by pushing the shift key at same time as 6 key) to rewrite the question. The order of operations called GEMA (Grouping, Exponents, Multiplication and Division from left to right and then Addition and Subtraction from left to right) can only be done once the symbols are placed to match the textbook question. Math is a language and its symbols must be used in place of words to assure the correct result.

7 0
4 years ago
Read 2 more answers
Solve: x^2+7x-30=0<br> Please show all the steps, thank u
vodomira [7]

x^2+7x-30=0 Rewrite the expression.

x^2+10x-3x-30=0 Factor the expressions.

x(x+10)-3 (x+10)=0 Factor the expression.

(x+10)(x-3)=0 Separate into possible cases.

x+10=0

Solve.

x-3=0

x1= -10

x2= 3

8 0
3 years ago
I NEED HELP PLS WHOEVER CAN ANSWER THIS CORRECTLY I WILL MAKE THEM THE BRAINLIEST
abruzzese [7]

Answer:

c.

Step-by-step explanation:

12-7=5

6-1=5

4 0
3 years ago
Read 2 more answers
is picking out some movies to rent, primarily interested in horror films and comedies. He has narrowed down his selections to 15
tigry1 [53]

Answer: 31,365

Step-by-step explanation:

Given : The number of horror films = 15

The number of comedies = 18

Then , the number of different combinations of 4 movies can he rent if he wants at least two comedies is given by :-

^{18}C_2\times ^{15}C_2+^{18}C_3\times ^{15}C_1+^{18}C_4\times ^{15}C_0\\\\=\dfrac{18!}{2!(18-2)!}\times\dfrac{15!}{2!(15-2)!}+\dfrac{18!}{3!(18-3)!}\times\dfrac{15!}{1!(15-1)!}+\dfrac{18!}{4!(18-4)!}\times\dfrac{15!}{0!(15-0)!}\\\\=16065+12240+3060=31365

Hence, the  number of different combinations of 4 movies can he rent if he wants at least two comedies = 31,365

7 0
3 years ago
CAN SOMEONE HELP ME WITH THIS?!?!
Nat2105 [25]
.......................:)
7 0
3 years ago
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