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4 years ago

9

What would be a good sentence for 36÷4

2 answers:

Aleks [24]4 years ago

8 0

36 divided by 4, hope this helps

Fofino [41]4 years ago

5 0

Thirty six Divided by four

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(-2.2); perpendicular to y= -6/5x +2

11111nata11111 [884]

(6.2) and the perpendicularly to y = to 1223

6 0

3 years ago

Simplify the following ratio ( Hint find a common factor and divide) 4 cats : 24 dogs

DedPeter [7]

Answer:

1 cat : 6 dogs

Step-by-step explanation:

4/2 : 2

24/2 : 12

2/2: 1

12/2: 6

Please mark brainliest!

Have an amazing day!

5 0

3 years ago

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pls help - will give brainliest farmer is painting a new barn. He will need to calculate the surface area of the barn to purchas

Brut [27]

Answer:

m^2

Step-by-step explanation:

You only need the surface area, which means you will not use cubed (^3). But the surface area is two dimensions (length and width). SO you must use m^2.

6 0

3 years ago

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A prism with a base area of 8 cm² and a height of 6 cm is dilated by a factor of 5/4 .

Andrei [34K]

The volume of the original prism is calculated by multiplying the area of the base by the height,
volume = (8 cm²)(6 cm) = 48 cm³
By the word dilated we mean to say that the dimensions of the new prism is 5/4 times that of the original solid. If we let x be the volume of the new solid,
x / 48 cm³ = (5/4)³
The value of x from the equation above is equal to 93.75 cm³.

8 0

3 years ago

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Consider the following higher-order differential equation. d3u dt3 + d2u dt2 − 2u = 0 Find all the roots of the auxiliary equati

g100num [7]

Answer:

Therefore the auxiliary solution is

$y=C_1e^{t}+e^{-t}[C_2sin \ t+C_3 cos \ t]$

Step-by-step explanation:

To find auxiliary equation we have to put $u=e^{mt}$ in the given differential equation.

The degree of the differential equation is 3.

Therefore the number of root of the differential equation is 3.

Let $\lambda_1$ , $\lambda_2$ and $\lambda_3$ be three roots of the auxiliary equation.

If three roots are real and equal.

Then $y= e^{\lambda_1t} (C_1+C_2t+C_3t^2)$

If three roots are real and distinct.

Then $y=C_1e^{\lambda_1t}+C_2e^{\lambda_2t}+C_3e^{\lambda_3x}$

If two roots imaginary and one root real , $\lambda_1= a+ib \ and \ \lambda_2= a-ib$

Then $y=C_1e^{\lambda_3t}+e^{at}(C_2sin \ bt+C_3cos \ bt)$

Now, $u=e^{mt}$ , $u'=me^{mt},u"=m^2e^{mt} \ and \ u'"= m^3e^{mt}$

Given differential equation is

$\frac{d^3u}{dt^3} +\frac{d^2u}{dt^2}-2u=0$

The auxiliary equation is

$(m^3+m^2-2)e^{mt}=0$

$\Rightarrow (m^3+m^2-2)=0$

$\Rightarrow m^3-m^2+2m^2-2m+2m-2=0$

$\Rightarrow m^2(m-1)+2m(m-1)+2(m-1)=0$

$\Rightarrow (m-1)(m^2+2m+2)=0$

$\Rightarrow m= 1,\frac{-2\pm\sqrt{2^2-4.2.1} }{2.1}$

$\Rightarrow m= 1,\frac{-2\pm\sqrt{-4} }{2.1}$

$\Rightarrow m= 1,-1\pm i$

Here $\lambda_1= -1+i , \lambda_2=- 1-i \ and \ \lambda_3=1$

Therefore ,

$y=C_1e^{1.t}+e^{-1.t}[C_2sin \ (1.t)+C_3 cos \ (1.t)]$

$\Rightarrow y=C_1e^{t}+e^{-t}[C_2sin \ t+C_3 cos \ t]$

5 0

4 years ago