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Bas_tet [7]
3 years ago
7

True or False: The angle opposite a side length of 6 cm in a triangle is larger than an angle opposite a side length of 7 cm in

the same triangle.
Mathematics
2 answers:
zepelin [54]3 years ago
5 0

Answer:

False

Step-by-step explanation:

it is not true!

kipiarov [429]3 years ago
3 0

Answer:

False

Step-by-step explanation:

The angle opposite the longest side is the largest

The angle opposite the shortest side is the smallest

So the angle opposite 6 cm is smaller than the angle opposite 7 cm

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Which statement describes f(x)=
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Answer:

A

Step-by-step explanation:

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2 years ago
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3 times (-8) times (-2)
dexar [7]

Answer:

48

Step-by-step explanation:

Signals during a multiplication of two numbers:

They have the same signal: The multiplication is positive.

They have different signals: The multiplication is negative.

3 times (-8) times (-2)​

First 3*(-8):

Different signals, so negative. 3*8 = 24. So the answer is -24.

Then:

3*(-8)*(-2) = (-24)*(-2)

Same signal, so positive. 24*2 = 48. The answer is 48.

4 0
1 year ago
Stacey filled her ½ cup measuring cup seven times to have enough flour for a cake recipe. How much flour does the cake recipe ca
alisha [4.7K]
Ok so in decimal form it would be 3.5
3 0
3 years ago
The logistic equation for the population​ (in thousands) of a certain species is given by:
Eva8 [605]

Answer:

a.

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
What is the number the difference of 11 and 14 times this number is equal to 115
Jet001 [13]

Answer:

-38.333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333...

Step-by-step explanation:

(11-14)×a=115

11-14=-3

-3×a=115

115÷-3=-38.33333333333333333333333333333333333333333...

5 0
3 years ago
Read 2 more answers
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