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2 years ago

13

5 Points

2 answers:

goldenfox [79]2 years ago

7 0

Answer:

320 ounces

Step-by-step explanation:

There is 16 ounces in one pound. Multiply 20 x 16 to get 320.

N76 [4]2 years ago

5 0

Answer:

the answer is 320 ounces

Step-by-step explanation:

just multiply 20 by 16 to find your answer

hope this helps! :)

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A) Find a particular solution to y" + 2y = e^3 + x^3. b) Find the general solution.

Reptile [31]

Answer:

a.P.I= $\frac{e^{3x}}{11}+\frac{1}{2}(x^3-3x)$

b.G.S= $C_1Cos \sqrt2 x+C_2 Sin\sqrt2 x+\frac{1}{11}e^{3x}+\frac{1}{2}(x^3-3x}$

Step-by-step explanation:

We are given that a linear differential equation

$y''+2y=e^{3x}+x^3$

We have to find the particular solution

P.I= $\frac{e^{3x}}{D^2+2}+\frac{x^3}{D^2+2}$

P.I= $\frac{e^{3x}}{3^2+2}+\frac{1}{2} x^3(1+\frac{D^2}{2})^{-2}$

P.I= $\frac{e^{3x}}{11}+\frac{1-2\frac{D^2}{4}+3\frac{D^4}{16}+...}{2}x^3$

P.I= $\frac{e^{3x}}{11}+\frac{x^3-2\frac{\cdot3\cdot 2x}{4}}{2}+0}$ (higher order terms can be neglected

P.I= $\frac{e^{3x}}{11}+\frac{1}{2}(x^3-3x)$

b.Characteristics equation

$D^2+2=0$

$D=\pm\sqrt2 i$

C.F= $C_1cos \sqrt2x+C_2 sin\sqrt2 x$

G.S=C.F+P.I

G.S= $C_1Cos \sqrt2 x+C_2 Sin\sqrt2 x+\frac{1}{11}e^{3x}+\frac{1}{2}(x^3-3x)$

3 0

3 years ago

Can someone please help me!!!!!!

Anna11 [10]

Answer:

It should be 1300..........

6 0

2 years ago

Read 2 more answers

What is the slope of the graph of y=-3X?<br><br> 3<br> 1<br> 3<br> 1<br> -3

Reil [10]

Answer:

-3

Step-by-step explanation:

y = -3x

This is in slope intercept form

y = mx+b where m is the slope and b is the y intercept

The slope is -3 and the y intercept is 0

5 0

1 year ago

Read 2 more answers

Noahzimmerman0987654321

sattari [20]

Answer:

ummm what is this for tho?

5 0

3 years ago

Triangles P Q R and S T U are shown. Angles P R Q and T S U are right angles. The length of P Q is 20, the length of Q R is 16,

dimaraw [331]

Answer:

$\angle P$

Step-by-step explanation:

Given

$\triangle PRQ = \triangle TSU = 90^o$

$PQ = 20$ $QR = 16$ $PR = 12$

$ST = 30$ $TU = 34$ $SU = 16$

<em>See attachment</em>

Required

Which sine of angle is equivalent to $\frac{4}{5}$

Considering $\triangle PQR$

We have:

$\sin(P) = \frac{QR}{PQ}$ --- i.e. opposite/hypotenuse

So, we have:

$\sin(P) = \frac{16}{20}$

Divide by 4

$\sin(P) = \frac{4}{5}$

Hence:

$\angle P$ is correct

7 0

2 years ago