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JulsSmile [24]
3 years ago
5

Light bulbs cost £2.90 each.

Mathematics
2 answers:
Natasha2012 [34]3 years ago
8 0

Answer:

<em>$2.60</em>

Step-by-step explanation:

s344n2d4d5 [400]3 years ago
7 0

Answer:

$2.60

Step-by-step explanation:

first you need to see how many time 2.90 goes into 20 it is 6, so then you multiply 2.90 times 6 and get 17.40. Lastly you subtract 17.40 from 20 to get 2.60

Hope this helps

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The function f(x) = 2x + 210 represents the number of calories burned when exercising, where x is the number of hours spent exer
docker41 [41]

Hi,

f + g is a simple sum of the functions. This refers to increasing the good effect of exercise with diet effect, therefore more weight could be lost.

(f + g)(x) = f(x) + g(x) = 2x + 210 + 2x + 125 = 4x + 335, so (f + g)(x) = 4x + 335.

For x = 3 (meaning 3 hours of exercise) we have:

(f + g)(3) = 4·3 + 335 = 12 + 335 = 347 calories, after 3 hours of exercice.

The correct answer is: 347 calories burned while combining diet with 3 hours of exercise.

Have you understood ?

Green eyes.

3 0
3 years ago
Alliya spent $6.70 on 12 pencils at the store. Her cost included $0.46 sales tax. How much did each pencil cost before the sales
Kaylis [27]
52 cents would be the answer.

Hope this helped. Would you like me to explain how I got it?
6 0
3 years ago
Read 2 more answers
Express 5601 in scientific notation
Anna71 [15]
In scientific notation it is:5000+600+1
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3 years ago
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
I don’t understand any of the below. Can you please help?
ehidna [41]
-25+(-12)=-37
14-(-20)=34
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6 0
2 years ago
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