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skad [1K]
3 years ago
12

A rectangular prism with integer side lengths has two faces with areas 15 and 21. What could its volume be?​

Mathematics
1 answer:
kari74 [83]3 years ago
5 0

Answer:

llllll

Step-by-step explanation:

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1/3(3+2x)−1=10<br><br><br> PLEASE HURRY I NEED It IN LIKE 2 MINUTES
Sholpan [36]

Answer:

x=15

Step-by-step explanation:

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3 years ago
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For every 11 litres of petrol a car can run for 50 km.
algol [13]

Answer: 50.2 liters

Step-by-step explanation:

11 liters = 50km

Z liters = 228km

To get the value of Z, cross multiply

228 x 11 = Z x 50

2508 = 50z

Divide both sides by 50

2508/50 = 50z/50

50.16 = z

50.16 has two decimal places, so convert to 1 decimal place by approximation

50.16 = 50.2

Thus, 50.2 liters will be enough for 228km

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3 years ago
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PLEASEE HELP ASAPPPP!<br> x9/x4 <br> PLEASSEEEEEEEE
Semmy [17]

Answer:

x^5

Step-by-step explanation

If this is dividing exponents, you just subtract the denominator from the numerator.

5 0
3 years ago
Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x​
Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

<em>r</em> ² - 3<em>r</em> + 2 = (<em>r</em> - 2) (<em>r</em> - 1) = 0

with roots <em>r</em> = 2 and <em>r</em> = 1, so the characteristic solution is

<em>y</em> (char.) = <em>C₁</em> exp(2<em>x</em>) + <em>C₂</em> exp(<em>x</em>)

For the <em>ansatz</em> particular solution, we might first try

<em>y</em> (part.) = (<em>ax</em> + <em>b</em>) + (<em>cx</em> + <em>d</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)

where <em>ax</em> + <em>b</em> corresponds to the 2<em>x</em> term on the right side, (<em>cx</em> + <em>d</em>) exp(<em>x</em>) corresponds to (1 + 2<em>x</em>) exp(<em>x</em>), and <em>e</em> exp(3<em>x</em>) corresponds to 4 exp(3<em>x</em>).

However, exp(<em>x</em>) is already accounted for in the characteristic solution, we multiply the second group by <em>x</em> :

<em>y</em> (part.) = (<em>ax</em> + <em>b</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)

Now take the derivatives of <em>y</em> (part.), substitute them into the DE, and solve for the coefficients.

<em>y'</em> (part.) = <em>a</em> + (2<em>cx</em> + <em>d</em>) exp(<em>x</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)

… = <em>a</em> + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)

<em>y''</em> (part.) = (2<em>cx</em> + 2<em>c</em> + <em>d</em>) exp(<em>x</em>) + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

… = (<em>cx</em> ² + (4<em>c</em> + <em>d</em>)<em>x</em> + 2<em>c</em> + 2<em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

Substituting every relevant expression and simplifying reduces the equation to

(<em>cx</em> ² + (4<em>c</em> + <em>d</em>)<em>x</em> + 2<em>c</em> + 2<em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

… - 3 [<em>a</em> + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)]

… +2 [(<em>ax</em> + <em>b</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)]

= 2<em>x</em> + (1 + 2<em>x</em>) exp(<em>x</em>) + 4 exp(3<em>x</em>)

… … …

2<em>ax</em> - 3<em>a</em> + 2<em>b</em> + (-2<em>cx</em> + 2<em>c</em> - <em>d</em>) exp(<em>x</em>) + 2<em>e</em> exp(3<em>x</em>)

= 2<em>x</em> + (1 + 2<em>x</em>) exp(<em>x</em>) + 4 exp(3<em>x</em>)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

<em>x</em> : 2<em>a</em> = 2

1 : -3<em>a</em> + 2<em>b</em> = 0

exp(<em>x</em>) : 2<em>c</em> - <em>d</em> = 1

<em>x</em> exp(<em>x</em>) : -2<em>c</em> = 2

exp(3<em>x</em>) : 2<em>e</em> = 4

Solving the system gives

<em>a</em> = 1, <em>b</em> = 3/2, <em>c</em> = -1, <em>d</em> = -3, <em>e</em> = 2

Then the general solution to the DE is

<em>y(x)</em> = <em>C₁</em> exp(2<em>x</em>) + <em>C₂</em> exp(<em>x</em>) + <em>x</em> + 3/2 - (<em>x</em> ² + 3<em>x</em>) exp(<em>x</em>) + 2 exp(3<em>x</em>)

4 0
3 years ago
Mariana solved the equation below. Identify her ERROR and EXPLAIN what she did wrong. -3(2x-4)=-12 -6x-12=-12 -6x=0 x=0
nydimaria [60]

Answer:

Her error was multiplying -3 and -4 which gives +12 but she wrote -12 instead

Step-by-step explanation:

Mariana workings:

-3(2x - 4) = -12

-6x - 12 = -12

Her error was multiplying -3 and -4 which gives +12 but she wrote -12 instead

-6x = 0

x=0

Correct workings:

-3(2x - 4) = -12

-6x + 12 = - 12

-6x = -12 - 12

-6x = -24

x = -24/-6

x = 4

8 0
3 years ago
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