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3 years ago

What is 8/30 in simplest form?

Mathematics

2 answers:

Free_Kalibri [48]3 years ago

5 0

8/30
= (8/2) / (30/2)
= 4/15

8/30 in its simplest form is 4/15~

Stella [2.4K]3 years ago

5 0

Check and see which number would go into both of them.
2 goes into both so 8 divided by 2 would be 4 and 30 divided by 2 would be 15
the solution is 4/15
hope this helps

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When cai traveled from new orleans to the ozark mountains in arkansas the elevation changed from 7ft below sea level to 2,314 ft

Tanya [424]

Answer:

2321 ft.

Step-by-step explanation:

When Cai traveled from New Orleans to the Ozark Mountains in Arkansas the elevation changes from 7 ft below sea level to 2314 ft above sea level.

If we consider the elevation below sea level as negative and the elevation above sea level as positive, then the elevation changes from - 7ft to +2314 ft.

Therefore, the increment of elevation is by [2314 - (- 7)] = 2321 ft. (Answer)

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3 years ago

If the right angle triangle LMN. L=30°, MN = 4cm and diagonal LM. Find the LM and LN.

Gnoma [55]

LM is 6.9 to 1dm and LM is 8

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3 years ago

Find the absolute maximum and minimum values of f(x, y) = x+y+ p 1 − x 2 − y 2 on the quarter disc {(x, y) | x ≥ 0, y ≥ 0, x2 +

Andreas93 [3]

Answer:

absolute max: f(x,y)=1/2+p1 ; at P(1/2,1/2)

absolute min: f(x,y)=p1 ; at U(0,0), V(1,0) and W(0,1)

Step-by-step explanation:

In order to find the absolute max and min, we need to analyse the region inside the quarter disc and the region at the limit of the disc:

Region inside the quarter disc:

There could be Minimums and Maximums, if:

∇f(x,y)=(0,0) (gradient)

we develop:

(1-2x, 1-2y)=(0,0)

x=1/2

y=1/2

Critic point P(1/2,1/2) is inside the quarter disc.

f(P)=1/2+1/2+p1-1/4-1/4=1/2+p1

f(0,0)=p1

We see that:

f(P)>f(0,0), then P(1/2,1/2) is a maximum relative

Region at the limit of the disc:

We use the Method of Lagrange Multipliers, when we need to find a max o min from a f(x,y) subject to a constraint g(x,y); g(x,y)=K (constant). In our case the constraint are the curves of the quarter disc:

g1(x, y)=x^2+y^2=1

g2(x, y)=x=0

g3(x, y)=y=0

We can obtain the critical points (maximums and minimums) subject to the constraint by solving the system of equations:

∇f(x,y)=λ∇g(x,y) ; (gradient)

g(x,y)=K

Analyse in g2:

x=0;

1-2y=0;

y=1/2

Q(0,1/2) critical point

f(Q)=1/4+p1

We do the same reflexion as for P. Q is a maximum relative

Analyse in g3:

y=0;

1-2x=0;

x=1/2

R(1/2,0) critical point

f(R)=1/4+p1

We do the same reflexion as for P. R is a maximum relative

Analyse in g1:

(1-2x, 1-2y)=λ(2x,2y)

x^2+y^2=1

Developing:

x=1/(2λ+2)

y=1/(2λ+2)

x^2+y^2=1

So:

(1/(2λ+2))^2+(1/(2λ+2))^2=1

$\lambda_{1}=\sqrt{1/2}*-1 =-0.29$

$\lambda_{2}=-\sqrt{1/2}*-1 =-1.71$

$\lambda_{2}$ give us (x,y) values negatives, outside the region, so we do not take it in account

For $\lambda_{1}$ : S(x,y)=(0.70, 070)

and

f(S)=0.70+0.70+p1-0.70^2-0.70^2=0.42+p1

We do the same reflexion as for P. S is a maximum relative

Points limits between g1, g2 y g3

we need also to analyse the points limits between g1, g2 y g3, that means U(0,0), V(1,0), W(0,1)

f(U)=p1

f(V)=p1

f(W)=p1

We can see that this 3 points are minimums relatives.

Conclusion:

We compare all the critical points P,Q,R,S,T,U,V,W an their respective values f(x,y). We find that:

absolute max: f(x,y)=1/2+p1 ; at P(1/2,1/2)

absolute min: f(x,y)=p1 ; at U(0,0), V(1,0) and W(0,1)

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3 years ago

The fastest race horse ran 1/4 mile in 1/3 of a minute. What was the horse's running speed?

bagirrra123 [75]

Answer:

3/4 miles per minute

Step-by-step explanation:

To get speed in miles per minute, just divide distance/time

1/4 mile / 1/3 minute = 3/4 miles per minute

If you want miles per hour, you'll have to convert minutes to hours.

3/4 miles/minute * 60 minutes/hour = 45 miles/hour

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3 years ago

Find the area of the circle to the nearest whole number, if necessary!

astra-53 [7]

Answer:

The circle has an area of about 1385 square mm.

Step-by-step explanation:

Let's recall that circles have an area that can be found with the following formula:

$A=\pi r^{2}$

where r is the radius of the circle.

Now, focus your eyes on the circle. We are shown that the diameter of this circle is 42 mm, but we only want the radius. Since the radius is half the diameter, the radius is 21 mm. Now, we can solve for the area of the circle.

$A=\pi (21)^2\\A=441\pi \\A=1385.44236...$

So, to the nearest whole number, the area of the circle is 1385 square mm.

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2 years ago