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3 years ago

Convert 5 kilograms to grams

Mathematics

2 answers:

konstantin123 [22]3 years ago

7 0

5,000 grams
1 kilogram=1,000 grams

ss7ja [257]3 years ago

5 0

Answer:

5000

Step-by-step explanation:

hmu if you need more help! :D

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Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x) = 6x(1/3) + 3x(4/3). You must justi

stealth61 [152]

Applying our power rule gets us our first derivative,

$\rm f'(x)=6\frac13x^{-2/3}+3\cdot\frac43x^{1/3}$

simplifying a little bit,

$\rm f'(x)=2x^{-2/3}+4x^{1/3}$

looking for critical points,

$\rm 0=2x^{-2/3}+4x^{1/3}$

We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.

$0=2x^{-2/3}\left(1+2x\right)$

When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.

Then apply your Zero-Factor Property,

$\rm 0=2x^{-2/3}\qquad\qquad\qquad 0=(1+2x)$

and solve for x in each case to find your critical points.

Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.

Let's come back to this,

$\rm f'(x)=2x^{-2/3}+4x^{1/3}$

and take our second derivative.

$\rm f''(x)=-\frac43x^{-5/3}+\frac43x^{-2/3}$

Looking for inflection points,

$\rm 0=-\frac43x^{-5/3}+\frac43x^{-2/3}$

Again, pulling out the smaller power of x, and fractional part,

$\rm 0=-\frac43x^{-5/3}\left(1-x\right)$

And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.

Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.

8 0

3 years ago

Can someone help please

Ymorist [56]

Answer:it is 7 and 6 that is your answer if that does not work it is 12 and 13 ok.

8 0

3 years ago

Find the length of the third side. If necessary, write in simplest radical form.

Mazyrski [523]

Answer:

$\boxed {\boxed {\sf 8}}$

Step-by-step explanation:

This triangle has a small square, which represents a right angle. Therefore, we can use the Pythagorean Theorem.

$a^2+b^2=c^2$

Where a and b are the legs of the triangle and c is the hypotenuse.

In this triangle, 7 and √15 are the legs, because these sides make up the right angle. The unknown side is the hypotenuse, because it is opposite the right angle. So, we know two values:

$a= 7 \\b= \sqrt{15}$

Substitute these values into the formula.

$(7)^2+(\sqrt{15})^2=c^2$

Solve the exponents.

(7)²= 7*7=49

$49+ (\sqrt{15})^2=c^2$

(√15)²=√15*√15=15

$49+15=c^2$

Add.

$64=c^2$

Since we are solving for c, we must isolate the variable. It is being squared and the inverse of a square is the square root. Take the square root of both sides.

$\sqrt{64}=\sqrt{c^2} \\\sqrt{64}= c\\8=c$

The third side length is 8.

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3 years ago

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olga2289 [7]

So for example 11/3 is 3 2/3 because you divide 11 by 3 and how you do it is that 3 goes into 11 3 times so you would write three on top of the 11 and then subtract 11-9 which is 2 so it is 3 2/3

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3 years ago

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If m∠D + m∠E = 180°, then ∠D and ∠E are classified as what kind of angles?

Archy [21]

The other person is literally wrong. The answer is D. Supplementary because two angles that add up to 180 are supplementary angles.

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2 years ago