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3 years ago

What is a good way to ace a level 5-7 maths test? I need some good tactics...subjects in particular.

2 answers:

5 0

It's important to study and study. I recommend studying the homework or study sheet the day before the test. Most likely the teacher has added a similar problem on to the test. Add a friend to help you study. To make it fun, buy candy and quiz yourself or friends! If you get a problem right, them give yourself candy.

weqwewe [10]3 years ago

4 0

Study, Study, Study!
The ONLY way you can ace a test is to study.
You can make flash cards, play online games(RELATED TO SUBJECT), or keep reading the lesson!

You might be interested in

Use the permutation formula to solve a problem when n = 15 and r = 4.

barxatty [35]

$nPr = \frac{n!}{(n - r)!}$

$15P4 = \frac{15!}{(15 - 4)!} = \frac{15!}{11!} \\ = \frac{15 \times 14 \times 13 \times 12 \times 11!}{11!} \\ 15 \times 14 \times 13 \times 12 = 32760$

3 0

3 years ago

Read 2 more answers

PLEASE HELP ME!!!!! I HAVE MORE QUESTIONS COMING

Whitepunk [10]

Use tide over run to help you find the answer

Hope that helps!!

8 0

3 years ago

While traveling to Europe, Phelan exchanged 250 US dollars for euros. He spent 150 euros on his trip. After returning to the Uni

Virty [35]

Answer:

$44.70

Step-by-step explanation:

250÷1.3687

=182.655074158

182.655074158-150

=32.655074158

32.655074158×1.3687

=44.6950000001

5 0

3 years ago

The logistic equation for the population (in thousands) of a certain species is given by:

Eva8 [605]

Answer:

b. 1.5

c. 1.5

d. No

Step-by-step explanation:

a. First, let's solve the differential equation:

$\frac{dp}{dt} =3p-2p^2$

Divide both sides by $3p-2p^2$ and multiply both sides by dt:

$\frac{dp}{3p-2p^2}=dt$

Integrate both sides:

$\int\ \frac{1}{3p-2p^2} dp =\int\ dt$

Evaluate the integrals and simplify:

$p(t)=\frac{3e^{3t} }{C_1+2e^{3t}}$

Where C1 is an arbitrary constant

I sketched the direction field using a computer software. You can see it in the picture that I attached you.

b. First let's find the constant C1 for the initial condition given:

$p(0)=3=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-1$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } =\frac{3}{2} =1.5$

c. As we did before, let's find the constant C1 for the initial condition given:

$p(0)=0.8=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=1.75$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2+1.75e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } =\frac{3}{2} =1.5$

d. To figure out that, we need to do the same procedure as we did before. So, let's find the constant C1 for the initial condition given:

$p(0)=2=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-\frac{1}{2} =-0.5$

Can a population of 2000 ever decline to 800? well, let's find the limit of the function when it approaches to ∞:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-0.5e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } =\frac{3}{2} =1.5$

Therefore, a population of 2000 never will decline to 800.

6 0

3 years ago

Use the long division method to find the result when x^3+7x^2+12x+6x 3 +7x 2 +12x+6 is divided by x+1x+1. If there is a remainde

Aliun [14]

Answer:

By long division (x³ + 7·x² + 12·x + 6) ÷ (x + 1) gives the expression;

$x^2 + 5 \cdot x + 7 - \dfrac{1}{(x + 1)}$

Step-by-step explanation:

The polynomial that is to be divided by long division is x³ + 7·x² + 12·x + 6

The polynomial that divides the given polynomial is x + 1

Therefore, we have;

$\ \ \ \ \ \ \ \ \ \ \ \ x^2 + 5\cdot x + 7\\ (x + 1) \sqrt{x^3 + 7\cdot x^2 +12\cdot x + 6} \\\ {} \ {} \ {} \ \ {} \ {} \ {} \ {} \ {} \ {} \ {} \ \ x^3 + 2 \cdot x^2 \\\ \ \ \ {} \ \ {} \ {} \ {} \ \ {} \ {} \ \ \ {} \ {} \ {} \ \ {} \ {} \ {} \ \ 5 \cdot x^2 + 12\cdot x + 6\\ \ {} \ {} \ {} \ {} \ {} \ {} \ \ {} \ {} 5 \cdot x^2 + 5\cdot x\\\ 7\cdot x+6\\7\cdot x+7\\-1$

(x³ + 7·x² + 12·x + 6) ÷ (x + 1) = x² + 5·x + 7 Remainder -1

Expressing the result in the form $q(x) + \dfrac{r(x)}{b(x)}$ , we have;

$(x^3 + 7\cdot x^2 + 12 \cdot x + 6)\div (x + 1) = x^2 + 5 \cdot x + 7 - \dfrac{1}{(x + 1)}$

4 0

3 years ago