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amid [387]
3 years ago
10

What is the smallest 3-digit number that is divisible by 2, 3, 4, 5, and 6?

Mathematics
2 answers:
tino4ka555 [31]3 years ago
6 0

Answer:Lcm (4,6)=12

Step-by-step explanation:Smallest 3 digit no. =100

Divide 100 by 12

100=12*8 +4

That means 12*8 would be the largest multiple of 12 that is less than 100

So 12*9 would be the smallest multiple of 12 that is greater than 100

Therefore the answer is 12*9=108.

Was it helpful?

Rasek [7]3 years ago
5 0

60

60/2=30

60/3=20

60/4=15

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972.8 feet is one minute
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Use the graph to find the solution of the system of equations {y = x^2 + 6x + 8 {y = x + 4
Phantasy [73]
Essentially when people ask you find the solution to system of equation, there asking at what x value do these to graphs intersect. The easiest way to do this is to get a graphing calculator, or desmos and type in the equation and find where they intersect. Heck, even the question says to solve it with a graph, but I'll demonstrate it algebraically.

One way you can do this is set the equation equal to each other. This is because you want to know at what x-value has the same y-value. So we get:

x^2 + 6x + 8 = x + 4

We can then combine like terms, or move everything to one side. So we get:

x^2 + 5x + 4 = 0.

Then we can use the quadratic formula to solve for x.

x=(-5 +/- sqrt(5^2 - 4(1)(4)))/(2(1)

This simplifies into:

(-5 +/- 3)/2

Finally we add and subtract:

(-5 + 3)/2 = x = -1
(-5 - 3)/2 = x = -4

And our solution is x = -1, x = -4
8 0
3 years ago
Please please please help and solve this with steps, much help needed thank you :) 20 points for this!
leonid [27]

Answer:

y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}

(Please vote me Brainliest if this helped!)

Step-by-step explanation:

\frac{dy}{dx}y=x^3+2x-5x+3

\mathrm{First\:order\:separable\:Ordinary\:Differential\:Equation}

  • \mathrm{A\:first\:order\:separable\:ODE\:has\:the\:form\:of}\:N\left(y\right)\cdot y'=M\left(x\right)

1. \mathrm{Substitute\quad }\frac{dy}{dx}\mathrm{\:with\:}y'\:

y'\:y=x^3+2x-5x+3

2. \mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}

yy'\:=x^3-3x+3

  • N\left(y\right)\cdot y'\:=M\left(x\right)
  • N\left(y\right)=y,\:\quad M\left(x\right)=x^3-3x+3

3. \mathrm{Solve\:}\:yy'\:=x^3-3x+3:\quad \frac{y^2}{2}=\frac{x^4}{4}-\frac{3x^2}{2}+3x+c_1

4. \mathrm{Isolate}\:y:\quad y=\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}}

5. \mathrm{Simplify}

y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}

7 0
3 years ago
Read 2 more answers
Please help me out i will mark brainliest!
asambeis [7]

Answer:

y=-x+2

x+y=2

Step-by-step explanation:

hopefully this helps :)

8 0
3 years ago
Read 2 more answers
given sin A =3/-/34 and that angle S is in quadrant 1, find the exact value of tan A in simplest radical form using a ration
Vedmedyk [2.9K]

Answer:

\tan(A) = \frac{3}{5}

Step-by-step explanation:

Given

\sin(A) = \frac{3}{\sqrt {34}}

0 \le A \le 90 --- First Quadrant

Required

Find tan(A)

The sin of an angle is:

\tan(A) = \frac{Opposite}{Hypotenuse}

and

\sin(A) = \frac{3}{\sqrt {34}}

By comparison:

Opposite = 3

Hypotenuse = \sqrt{34

So, the Adjacent is:

Hypotenuse^2 = Adjacent^2 + Opposite^2

(\sqrt{34})^2 = Adjacent^2 + 3^2

34 = Adjacent^2 + 9

Collect like terms

Adjacent^2 =34 - 9

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Take square roots

Adjacent =\sqrt{25

Adjacent =5

The tangent of an angle is:

\tan(A) = \frac{Opposite}{Adjacent}

This gives:

\tan(A) = \frac{3}{5}

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