All categories

4 years ago

Represent in number line 1/5 and 2/5

Mathematics

2 answers:

vovangra [49]4 years ago

6 0

See picture for math work and answer.

sashaice [31]4 years ago

5 0

I’m not sure if I’m right but here’s my answer:

You might be interested in

Which of the following is a factor of xy + 3y − 7x − 21

NISA [10]

What are the options?

8 0

4 years ago

What does 3k-1=7k+ 2 equal?

PIT_PIT [208]

K= -3/4 would be your answer

4 0

4 years ago

Kayla is A high school basketball player. In a particular game, she made some two point shots and some three point shots. Kayla

Mrrafil [7]

Answer:

she made 5 two point shots and 4 three point shots

Step-by-step explanation:

5 x 2 = 10 and 4 x 3 = 12. 10 + 12 = 22

4 0

3 years ago

Read 2 more answers

Consider the differential equation,

kondor19780726 [428]

Answer:

The two solutions are given as $y_1(t)=\dfrac{1}{9}e^{-8t}+\dfrac{8}{9}e^{t}$ and $y_2(t)=\dfrac{-1}{9}e^{-8t}+\dfrac{1}{9}e^{t}$

Step-by-step explanation:

As the given equation is

$y''+7y'-8y=0\\$

So the corresponding equation is given as

$m^2+7m-8=0$

Solving this equation yields the value of m as

$(m+8)(m-1)=0\\m=-8, m=1$

Now the equation is given as

$y(t)=C_1e^{m_1t}+C_2e^{m_2t}$

Here m1=-8, m2=1 so

$y(t)=C_1e^{-8t}+C_2e^{t}$

The derivative is given as

$y'(t)=-8C_1e^{-8t}+C_2e^{t}$

Now for the first case y(t=0)=1, y'(t=0)=0

$y(t=0)=C_1e^{-8*0}+C_2e^{0}\\1=C_1+C_2\\\\y'(t=0)=-8C_1e^{-8*0}+C_2e^{0}\\0=-8C_1+C_2$

So the two equation of co-efficient are given as

$C_1+C_2=1\\-8C_1+C_2=0$

Solving the equation yield

$C_1=1/9 \\C_2=8/9$

So the function is given as

$y_1(t)=\dfrac{1}{9}e^{-8t}+\dfrac{8}{9}e^{t}$

Now for the second case y(t=0)=0, y'(t=0)=1

$y(t=0)=C_1e^{-8*0}+C_2e^{0}\\0=C_1+C_2\\\\y'(t=0)=-8C_1e^{-8*0}+C_2e^{0}\\1=-8C_1+C_2$

So the two equation of co-efficient are given as

$C_1+C_2=0\\-8C_1+C_2=1$

Solving the equation yield

$C_1=-1/9 \\C_2=1/9$

So the function is given as

$y_2(t)=\dfrac{-1}{9}e^{-8t}+\dfrac{1}{9}e^{t}$

So the two solutions are given as $y_1(t)=\dfrac{1}{9}e^{-8t}+\dfrac{8}{9}e^{t}$ and $y_2(t)=\dfrac{-1}{9}e^{-8t}+\dfrac{1}{9}e^{t}$

7 0

3 years ago

Can someone help me please?

katrin2010 [14]

Answer:

120,000-30,000=$90000 in bonds or $4500 profit.

Step-by-step explanation:

4 0

3 years ago