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2 years ago

Please help pleaseeee I really need answers only

Mathematics

2 answers:

stiks02 [169]2 years ago

7 0

☁️ Answer ☁️

5. IS E (none of these)

Because I got:

-125y^5/162x

6. X = 5/3

Hope it helps.

Have a nice day noona/hyung.

sasho [114]2 years ago

3 0

<u>Answer:</u>

You might be interested in

HELP ITS DUE TMMR !!!!

pentagon [3]

Answer:

a) B

b) draw a closed circle right before 0.1 (where 0.083 would be)

c) 1/12

6 0

2 years ago

Sofia has 25 coins in nickels and dimes in her pocket for a total of $1.65 how many of each type of coin does she have?

Simora [160]

Answer:

She has 23 nickels and 5 dimes.

Step-by-step explanation:

Let n = the number of nickles and d = the number of dimes.

The number of each = 28. Set up the equation:

n + d = 28

Solve for n:

n = 28 - d

The total amount is $1.65 or 165 cents:

5n + 10d = 165

Substitute:

5(28 - d) + 10d = 165

140 - 5d + 10d = 165

140 + 5d = 165

5d = 165 - 140

5d = 25

d = 5, the number of dimes.

Solve for n:

n = 28 - d

n = 28 - 5 = 23, the number of nickles.

Proof:

5n + 10d = 165

5(23) + 10(5) = 165

115 + 50 = 165

165 = 165

Hope this Helps! Have an Awesome Day!!! <3

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Lynne bought a bag of grapefruit, 1 5/8 pounds of apples, and 2 3/16 pounds of bananas.The total weight of her purchases was 7 1

Rufina [12.5K]

The bag of grapefruit weighs 3 and 11/16
first: find a common denominator of 1/2, 5/8, and 3/16; 16 is the least common multiple.
16 divided by 2 is 8, so we multiply 1 by 8 and have 8/16
16 divided by 8 is 2, so we multiply 5 by 2 and have 10/16.
now, the three quantities given are 1 and 10/16, 2 and 3/16, and 7 and 8/16.
then, we subtract 1 and 10/16 from 7 and 8/16
7 8/16
- 1 10/16
this is 5 and 7/8, or 5 and 14/16

then, we subtract 2 and 3/16 from 5 and 14/16 to get 3 and 11/16

there you have it :)

4 0

3 years ago

Please help me out ill mark u brainlist

True [87]

Answer:

a = 13

Step-by-step explanation:

Isolate the variables by dividing each side factors that do not contain the variable.

6 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