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3 years ago

How to solve (5 x 2)^10 -2/ 2^5

Mathematics

1 answer:

seraphim [82]3 years ago

8 0

Answer: 1: multiply the numbers

2: Evaluate the exponent

3: solution 159/16

Step-by-step explanation:

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2x+x(x+6)<br> Solve to the lowest equation/number

Ghella [55]

Answer:

x^2 +8x

Step-by-step explanation:

2x+x(x+6)

Distribute

2x + x^2 +6x

Combine like terms

x^2 +8x

7 0

3 years ago

Read 2 more answers

zzz [600]

Answer:1 computer : 8 students

Step-by-step explanation:

if there are 9 computers it would originally be 9 : 72 but that can be reduced since 9 is a factor of 72.

8 0

3 years ago

Read 2 more answers

(02.06)If 1 pint is equal to 2 cups, then 5 pints would equal ___ cups.

EleoNora [17]

Answer:

If 1 pint is equal to 2 cups, then 5 pints would equal _<u>5</u>_ cups.

Step-by-step explanation:

Let's use a ratio to solve:

pints : cups

1 : 2

2 : 4

3 : 6

4 : 8

5 : 10

There would be 10 cups in 5 pints.

3 0

3 years ago

There are 12 red marbles and 8 green marbles in a bag. What is the probability of selecting a red

madam [21]

Answer:

The probability of selecting a red marble, not replacing it, and then selecting a green marble from the bag is 24/95

Step-by-step explanation:

Number of red marbles = 12

Number of green marbles = 8

Total number of marbles = 12+8 = 20

Probability of selecting red marble = $\frac{12}{20}$

Since it is the case of no replacement

Remaining marbles = 20-1 = 19

Number of red marbles = 12-1=11

Number of green marbles = 8

Probability of selecting green marble = $\frac{8}{19}$

So, the probability of selecting a red marble, not replacing it, and then selecting a green marble from the bag = $\frac{12}{20} \times \frac{8}{19}=\frac{24}{95}$

Hence the probability of selecting a red marble, not replacing it, and then selecting a green marble from the bag is 24/95

3 0

4 years ago

Factor the quadratic equation below to reveal the solutions. X^2+4x-21=-9

a_sh-v [17]

Answer:

x = 3 and x = -7

Step-by-step explanation:

The given quadratic equation is $x^2+4x-21=0$ . We need to find the solution of this equation.

If the equation is in the form of $ax^2+bx+c=0$ , then its solutions are given by :

$x=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}$

Here, a = 1, b = 4 and c = -21

Plugging all the values in the value of x, such that :

$x=\dfrac{-b+ \sqrt{b^2-4ac} }{2a},\dfrac{-b- \sqrt{b^2-4ac} }{2a}\\\\x=\dfrac{-4+ \sqrt{(4)^2-4\times 1\times (-21)} }{2(1)},\dfrac{-4- \sqrt{(4)^2-4\times 1\times (-21)} }{2(1)}\\\\x=3, -7$

So, the solutions of the quadratic equation are 3 and -7.

6 0

3 years ago