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lesya [120]
2 years ago
11

The base of this prism is______________ triangle.

Mathematics
2 answers:
Elena L [17]2 years ago
3 0

Answer:

b

Step-by-step explanation:

scoray [572]2 years ago
3 0

Answer:

b an isosceles

Step-by-step explanation:

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Ronnie collects horseshoes and hand grenades and places
lesantik [10]

Answer: 3.5 pounds

Step-by-step explanation:

3 s   + 7 h = 32        s = shoes   h= hand grenades

6 s   + 2 h = 22        Multiply first equation by -2   and add to second equation to get :

-12 h = -42

h = 3.5 pounds

sorry if this was late

8 0
2 years ago
Simplify 4^7/5^2 all raised to the 3rd power<br><br> 4^10/5^5<br> 4^4/5<br> 4^21/5^6<br> 12^7/15^2
zepelin [54]
[4^7/5^2]³ = 4^21/5^6

When a power is raised to an other power, you multiply these 2 powers

Answer: 4^21/5^6
5 0
3 years ago
PLEASE ANSWER WILL MARK BRAINLIEST
Anastasy [175]

Answer: the 2nd bubble B

Step-by-step explanation:

4 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
Monica is considering a 7/23 balloon mortgage with an interest rate of 4.5% to purchase a house for $204,000. What will be her b
arlik [135]

Answer:

$177,533.62

Step-by-step explanation:

AP3X

5 0
3 years ago
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