F(x)=2x-1, g(x)= 2-x^2 find g(-1)

Mathematics

1 answer:

rjkz [21]3 years ago

5 0

Answer:

Step-by-step explanation:

g(x)=2-x^2

g(-1)=2-(-1)^2

=2-1

Hope that this is helpful.

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90 points****

krek1111 [17]

Answer:

The answer would be the second choice <em><u>3.78 m²</u></em>

Step-by-step explanation:

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<em>Hope this Helps!</em>

3 0

3 years ago

What is the HCF of 36,60,84?

eimsori [14]

<h3><em>(HCF)=(36,60,84)</em></h3><h3><em>Factors of 36 </em></h3><h3><em> </em></h3><h3><em>List of positive integer factors of 36 that divides 36 without a remainder. </em></h3><h3><em> </em></h3><h3><em>1,2,3,4,6,9,12,18,36</em></h3><h3>Factors of 60 </h3><h3> </h3><h3>List of positive integer factors of 60 that divides 60 without a remainder. </h3><h3> </h3><h3>1,2,3,4,5,6,10,12,15,20,30,60</h3><h3>Factors of 84 </h3><h3> </h3><h3>List of positive integer factors of 84 that divides 84 without a remainder. </h3><h3> </h3><h3>1,2,3,4,6,7,12,14,21,28,42,84</h3><h3 /><h3>We found the factors 36,60,84 . The biggest common factor number is the HCF number. </h3><h3>So the highest common factor 36,60,84 is 12.</h3><h3><em>HOPE IT HELPS....</em></h3>

5 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of f(x,y)=x+4y subject to the constraint x2+y2=9, if such values

Vesnalui [34]

The Lagrangian is

$L(x,y,\lambda)=x+4y+\lambda(x^2+y^2-9)$

with critical points where the partial derivatives vanish.

$L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}$

$L_y=4+2\lambda y=0\implies y=-\dfrac2\lambda$

$L_\lambda=x^2+y^2-9=0$

Substitute $x,y$ into the last equation and solve for $\lambda$ :

$\left(-\dfrac1{2\lambda}\right)^2+\left(-\dfrac2\lambda\right)^2=9\implies\lambda=\pm\dfrac{\sqrt{17}}6$

Then we get two critical points,

$(x,y)=\left(-\dfrac3{\sqrt{17}},-\dfrac{12}{\sqrt{17}}\right)\text{ and }(x,y)=\left(\dfrac3{\sqrt{17}},\dfrac{12}{\sqrt{17}}\right)$

We get an absolute maximum of $3\sqrt{17}\approx12.369$ at the second point, and an absolute minimum of $-3\sqrt{17}\approx-12.369$ at the first point.