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3 years ago

Translate please The quotient of a number and the sum of the number and 5 is -2

Mathematics

1 answer:

julia-pushkina [17]3 years ago

6 0

X/X+5=-2
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What is (x^-2 y^10)^2/x^5 y^-3 using only positive exponents

Eddi Din [679]

Answer:

$\frac{(x^{ - 2}{y}^{10})^{2} }{ {x}^{5} {y}^{ - 3} } \\ \frac{ {x}^{ - 4} {y}^{20} }{ {x}^{5} {y}^{ - 3 } } \\ {x}^{ - 9} {y}^{23} \\ \frac{ {y}^{23} }{ {x}^{9} } \\ ans \: c$

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1 year ago

Jonathan bought a calculator for X$. Cindy bought the same calculator for 3 times more than Jonathan paid for his calculator. Wh

Andrej [43]

Answer:

3x or 3*x

Step-by-step explanation:

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2 years ago

What is the answer when you evaluate n + m -2(m-m) when m = 3 and n = 2?

Effectus [21]

Answer:

Step-by-step explanation:

n + m -2(m-m)

Combine like terms

n+m+2(0)

n+m

Let m=3 and n=2

2+3

7 0

2 years ago

ſ x - 5y = -9( 4x + 4y = - 12Step 1 of 2: Determine if the point (-4,1) lies on both of the lines in the system of equations by

Phoenix [80]

Given:

$\begin{gathered} x-5y=-9\ldots\ldots(1) \\ 4x+4y=-12\ldots\text{.}\mathrm{}(2) \end{gathered}$

The given point is (-4,1)

Let's check the ordered pair (-4,1) in the first equation

$\begin{gathered} x-5y=-9 \\ (-4)-5(1)=-9 \\ -4-5=-9 \\ -9=-9 \end{gathered}$

Hence, (-4,1) is a solution of the first equation x-5y=-9.

Now, let's check (-4,1) in the second equation.

$\begin{gathered} 4x+4y=-12 \\ 4(-4)+4(1)=-12 \\ -16+4=-12 \\ -12=-12 \end{gathered}$

So, (-4,1) is a solution of the second equation 4x+4y=-12.

Hence, (-4,1) is a solution of both equation in the system, then it is a solution to the overall system.

3 0

2 months ago

If the initial amount of iodine-131 is 537 grams , how much is left after 10 days?

viktelen [127]

Answer:

225.78 grams

Step-by-step explanation:

To solve this question, we would be using the formula

P(t) = Po × 2^t/n

Where P(t) = Remaining amount after r hours

Po = Initial amount

t = Time

In the question,

Where P(t) = Remaining amount after r hours = unknown

Po = Initial amount = 537

t = Time = 10 days

P(t) = 537 × 2^(10/)

P(t) = 225.78 grams

Therefore, the amount of iodine-131 left after 10 days = 225.78 grams

4 0

2 years ago