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3 years ago

Using capital letters and no periods or spaces, type the abbreviation for the congruence postulate.

Mathematics

2 answers:

worty [1.4K]3 years ago

8 0

Answer:

Step-by-step explanation:

jek_recluse [69]3 years ago

6 0

HL is the answer got it from answer key

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Pealse help meeeeee I NEED HELP

nalin [4]

Answer:

Step-by-step explanation:

1 a function states that for every y coordinate there is on x coordinate

2 the y intercept is -1 slope is 0

3 i do not know im trying to figure out dont delete i will find it

6 0

3 years ago

What is the area of the triangle?

Strike441 [17]

Answer:

Step-by-step explanation:

base (b) = 8

height (h) = 6

A = bh/2

= 8x6/2

= 48/2

= 24

Therefore the area is 24 square units

3 0

3 years ago

Question 67 and 68 please. Or one or the other.

Umnica [9.8K]

Answer:

question 67-the answer is Rory has eaten 1 1/4 of all the chocolate he has been given.

question 68-answer A-30 degrees

Step-by-step explanation:

question 67:

first Rory ate 1/2 of his chocolate

then, his sister gave him another whole chocolate bar an hour later.

that is 1/2+1=1 1/2--3/2/2=3/4

add 3/4 and 1/2--3/4 1/2~2/4--2/4+3/4=5/4 or 1 1/4

so the answer is 1 1/4

question 68:

if you draw a base to connect the last side of the angle into a triangle, you find that all three sides are equal, making the triangle an equilateral triangle. An equilateral triangle's angles all equal 30 degrees.

so, the answer is answer A: 30 degrees

6 0

3 years ago

How do I do this plz help me

USPshnik [31]

Naming the circle is just the only point in the center of the circle (E). The radius is half diameter of the circle so imagine a line connecting the center dot to any dot to the outside. An example of this would be (EC). The diameter is half the circle so any 3 points that cut the circle in half. This could be either (DEC) or (AEB).
So the answers would be:
A: (E)
B: (EC)
C: (DEC)
D: (AEB)

3 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