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2 years ago

NO NEED FOR LINKS NEED HELP RIGHT NOW

Mathematics

2 answers:

SVEN [57.7K]2 years ago

8 0

1 m = 0.001 Km
1 Km = 1000 m
1 ml = 0.001 L
1 L = 1000 ml
1 g = 0.001 Kg
1 Kg = 1000 g
1 mm = 0.01 cm

a. 2000 m = 2 Km b. 9 Km = 9000m
a. 9000 ml = 9 L b. 3 Kg = 3000 g
a. 6 L = 6000 ml b. 90 mm = 0.9 cm

Hope you could get an idea from here.

Doubt clarification - use comment section.

Softa [21]2 years ago

4 0

Answer:

1a.2

2a. 9

3a. 6000

1b. 9000

2b. 3000

3b. 9

Step-by-step explanation:

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Simplify the polynomial 3x^2+5x-5x^2-4x+5-2

zepelin [54]

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3 years ago

Fill in the blanks to make true number sentences.<br><br> 1 3/5 = __ : __

arsen [322]

Answer:

1 6/10

Step-by-step explanation:

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3 years ago

Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview

mario62 [17]

Answer: The required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.$

So, the general solution of the given equation is

$y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.$

According to the given conditions, we have

$y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

and

$y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.$