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tatuchka [14]
3 years ago
8

x - y) { }^{2} " align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Katyanochek1 [597]3 years ago
4 0
(x-y)^2 is equivalent to 

x^2 - 2xy + y^2.
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beth is planning a playground and has decided to place the swings in such a way that they are thes same distance from the jungle
NemiM [27]

Answer:

i dont know

Step-by-step explanation:

8 0
3 years ago
Can someone please help me?
sasho [114]

Answer:

C is correct because if she pays 700 dollars for a car payment her amount of money to spent money ratio will be above the recommended amount of spent money percentage to amount of money ratio, therefore meaning she will be in credit overload.

Step-by-step explanation:

3 0
3 years ago
Element X is a radioactive isotope such that its mass decreases by 26% every day. If an experiment starts out with 810 grams of
ioda

Answer:

The hourly decay rate is of 1.25%, so the hourly rate of change is of -1.25%.

The function to represent the mass of the sample after t days is A(t) = 810(0.74)^t

Step-by-step explanation:

Exponential equation of decay:

The exponential equation for the amount of a substance is given by:

A(t) = A(0)(1-r)^t

In which A(0) is the initial amount and r is the decay rate, as a decimal.

Hourly rate of change:

Decreases 26% by day. A day has 24 hours. This means that A(24) = (1-0.26)A(0) = 0.74A(0); We use this to find r.

A(t) = A(0)(1-r)^t

0.74A(0) = A(0)(1-r)^{24}

(1-r)^{24} = 0.74

\sqrt[24]{(1-r)^{24}} = \sqrt[24]{0.74}

1 - r = (0.74)^{\frac{1}{24}}

1 - r = 0.9875

r = 1 - 0.9875 = 0.0125

The hourly decay rate is of 1.25%, so the hourly rate of change is of -1.25%.

Starts out with 810 grams of Element X

This means that A(0) = 810

Element X is a radioactive isotope such that its mass decreases by 26% every day.

This means that we use, for this equation, r = 0.26.

The equation is:

A(t) = A(0)(1-r)^t

A(t) = 810(1 - 0.26)^t

A(t) = 810(0.74)^t

The function to represent the mass of the sample after t days is A(t) = 810(0.74)^t

5 0
3 years ago
NEED HELP WITH THESE QUESTIONS
amm1812

For this case we must solve the following questions:

Question 1:

We should simplify the following expression:

\frac {\frac {m ^ 2 * n ^ 3} {p ^ 3}} {\frac {mp} {n ^ 2}} =

Applying double C we have:

\frac {m ^ 2 * n ^ 3 * n ^ 2} {mp * p ^ 3} =

By definition of multiplication of powers of the same base we have to place the same base and add the exponents:\frac {m ^ 2 * n ^ 5} {m * p ^ 4} =

Canceling common terms:

\frac {mn ^ 5} {p ^ 4}

Answer:

Option A

Question 2:

We should simplify the following expression:

\frac {3xyz ^ 2} {6y ^ 4} * \frac {2y} {xz ^ 4}

So, we have:

\frac {3xyz ^ 2 * 2y} {6y ^ 4 * xz ^ 4} =\\\frac {6xy ^ 2z ^ 2} {6y ^ 4xz ^ 4} =

Simplifying common terms:

\frac {1} {y ^ 2z ^ 2}

Answer:

Option D

Question 3:

We factor the following expressions to rewrite the experience:

<em>r ^ 2 + 7r + 10: </em>We look for two numbers that multiplied give 10 and added 7:

(r + 5) (r + 2)

<em>r ^ 2-5r-50:</em> We look for two numbers that multiplied give -50 and added -5:

(r-10) (r + 5)

<em>3r-30 = 3 (r-10)</em>

Rewriting the given expression we have:

\frac {(r + 5) (r + 2) * 3 (r-10)} {3 (r-10) (r + 5)} =

We simplify common terms in the numerator and denominator we have:

(r + 2)

Answer:

Option D

4 0
3 years ago
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What is 3 to the power of 8 divided by 3 to the power of 2
Troyanec [42]

Answer:

729

Step-by-step explanation:

hope this helps!

3 0
3 years ago
Read 2 more answers
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