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3 years ago

If 20 is decreased by 40%, what is the new amount?

Mathematics

2 answers:

tester [92]3 years ago

8 0

Please go on my acc and help

aksik [14]3 years ago

3 0

Answer: The correct answer is 12.

Step-by-step explanation: If 20 is decreased by 40% the first thing that we need to do is calculate how much is 40% of 20.

20 x .4 = 8

Now, subtract 8 from 20:

The answer:

20 - 8 = 12

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Why does a logarithmic equation sometimes have an extraneous solution?

jonny [76]

The main reason behind this is using properties of logarithm .

when solving

log_2(x+2)+log_2(x-3)=3

There you use multiplication property and make addition to multiplication then you get extraneous solution because in plenty of cases they occur

(-2)²=(2)²

But

-2≠2

Same happens in case of logarithm

6 0

2 years ago

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Help plzzzz<br> I will mark you as a brainliest

Phoenix [80]

Substitute the x values back into the equation

(1, 22)
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3 years ago

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Exercice 2: Un remorqueur tire un bateau à la vitesse de 10 noeuds. La tension du câble est de 2500N, Le câble est parallèle à l

Elza [17]

Répondre:

1 285 watts

Explication étape par étape:

La puissance est exprimée selon la formule;

power = travail effectué / temps

Puissance = Force * (distance / temps)

Depuis Vitesse = distance / temps

Puissance = Force * vitesse

Donné

Force = 2500N

Vitesse (en m / s) = 0,514 m / s

Obligatoire

Puissance

Remplacez la formule donnée;

Puissance = 2500 * 0,514

Puissance = 1,285Watts

Par conséquent, la puissance correspondante requise est de 1285 watts

5 0

3 years ago

The steps for adding integers are the same as for subtracting them. <br><br> True or False?

lisov135 [29]

Answer:

I'm really going to say false im not sure, I just started learning this.

Step-by-step explanation:

4 0

3 years ago

Consider the functions below. f(x, y, z) = x i − z j + y k r(t) = 4t i + 6t j − t2 k (a) evaluate the line integral c f · dr, wh

fredd [130]

With

$\vec r(t)=4t\,\vec\imath+6t\,\vec\jmath-t^2\,\vec k$

we have

$\mathrm d\vec r=(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt$

The vector field evaluated over this parameterization is

$\vec f(x,y,z)=\vec f(x(t),y(t),z(t))=4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k$

so the line integral is

$\displaystyle\int_{-1}^1(4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k)\cdot(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt$

$=\displaystyle\int_{-1}^1(16t+6t^2-12t^2)\,\mathrm dt=-4$

6 0

3 years ago