MATHEMATICS COLLEGE

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Answer 1

Answer:

Answer:

4a x 2a + 0

Step-by-step explanation:

4a x 2a = 8a

0 + 0 = 0

= 8a, 0

Answer 2

Answer:

Answer:

brainlyist

Step-by-step explanation:

x=5 y=0

x=0 y=3

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Kyle has 64 baseball cards and must give away 1/4 of his figures away. How many cards will Kyle give away? What fractions of his cards does he have left?

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kyle regalara 16 figuras y le quedara 3/4 de ellas

If C=X + 6 and D=3X -6 + 4x to the second power of 4x if an expression that equals C + 3D in standard form.

Answers

Answer:

3=3y4/i94 =78

Step-by-step explanation:

Please answer fast will make brainlest

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Answer:

Step-by-step explanation:

C^2 + 5c =3 Tell if this factors or not? And show how it factors or does not factor? And if it factors what are the roots?

Answers

The answer to your questions is that it depends on how we view the polynomial. In particular,

$c^2+5c=3\implies c^2+5c-3=0$

If the left hand side were factorizable, then we would be able to write it in the form

$(c-r_1)(c-r_2)=c^2+5c-3=0$

If we expand the leftmost expression, we'd get

$c^2-(r_1+r_2)c+r_1r_2=c^2+5c-3=0$

and so for the two polynomials to be the same, the coefficients must match. In other words, the unknowns $r_1,r_2$ would have to satisfy

$\begin{cases}-(r_1+r_2)=5\nr_1r_2=-3\end{cases}$

Suppose that, moreover, we want integer solutions for $r_1,r_2$ . For this to happen, they must be factor pairs of the constant term.

-3 only has two factor pairs. Either $r_1=-1$ and $r_2=3$ , or $r_1=1$ and $r_2=-3$ . In the first case, we'd get a linear coefficient of $-(-1+3)=-2\neq5$ , while in the second, we'd get $-(1-3)=2\neq5$ .

There is no integer solution for this system, so the original quadratic is not factorizable - but only so over the integers.

If we change the scope of the coefficients, i.e. allow for any real numbers/complex numbers to appear in the factorization, then we always factorize a quadratic. The above system is easy to solve.

$r_1r_2=-3\implies r_2=-\frac3{r_1}$
$\implies-\left(r_1-\frac3{r_1}\right)=5$
$\implies{r_1}^2+5r_1-3=0$
$\implies r_1=\frac{-5\pm√(37)}2$

$\implies c^2+5c-3=\left(c+\frac{5-√(37)}2\right)\left(c+\frac{5+√(37)}2\right)$

so the original quadratic is factorizable over the reals.

On 1st September 2014 there were 5400 trees planted in a wood.On 1st September 2015, only 5184 of these trees were still alive.
It is assumed that the number of trees still alive is given by N = art
where / is the number of trees still alive t years after 1st September 2014.
a) Write down the value

c) Show that on 1st September 2040
the number of trees still alive is predicted
o have decreased by over 65% compared
with September 2014.

b) Show that r = 0.96

Answers

Answer:

1. a = 5400

2. r = 0.96

3. Percentage decrement = 65.4%

Step-by-step explanation:

Given

N = ar^t

Solving (a): Write down the value of a

a implies the first term

And from the question, we understand that the initial number of trees is 5400.

Hence,

a = 5400

Solving (b): Show that r = 0.96

Using

N = ar^t

When a = 5400, t = 1 i.e. the first year and N = 5184

Substitute these values in the above expression

5184 = 5400 * r¹

5184 = 5400 * r

5184 = 5400r

Solve for r

r = 5184/5400

r = 0.96

Solving (c): Show that the trees has decreased by over 65% in 2040

First, we need to calculate number of years (t) in 2040

t = 2040 - 2014

t = 26

Substitute 26 for t, 5400 for a and 0.96 for r in N = ar^t to get the number of trees left

N = 5400 * 0.96^26

N = 1868.29658019

N = 1868 (approximated)

Next, we calculate the percentage change as thus:

%Change = (Final - Initial)/Initial * 100%

Where the initial number of trees =5400 and final = 1868

%Change = (1868 - 5400)/5400 * 100%

%Change = -3532/5400 * 100%

%Change = -3532%/54

%Change = -65.4%

The negative sign indicates a decrements or reduction.

Hence, percentage decrement = 65.4% and this is over 65%

How are 0.5 and 0.05 and 0.0005 related

Answers

$\bf \cfrac{5}{10}\cdot \cfrac{1}{10}\implies \cfrac{5}{100}~\hspace{10em}\cfrac{5}{100}\cdot \cfrac{1}{10}\implies \cfrac{5}{1000}\n\n\n\textit{we can say one is \boxed{\textit{one tenth of the other}}}\n\n[-0.35em]~\dotfill\n\n\cfrac{5}{100}\cdot 10\implies \cfrac{5}{10}~\hspace{10em}\cfrac{5}{1000}\cdot 10\implies \cfrac{5}{100}\n\n\n\textit{or we can say, one is \boxed{\textit{ten times the other}}}$