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2 years ago

Simplify the following 2y^4 x 5y^3

Mathematics

2 answers:

sveta [45]2 years ago

5 0

Answer:

10 $y^{7}$

Step-by-step explanation:

2 $y^{4}$ · 5y³

10 $y^{3}+y^{4}$

10 $y^{7}$

adelina 88 [10]2 years ago

3 0

Answer:

10y^7

Step-by-step explanation:

because 2x5=10

and when multiplying with powers you add the powers together so 4+3=7

now you have 10y^7

hope this helped!!

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Janie is analyzing a quadratic function f(x) and a linear function g(x). Will they intersect?

UkoKoshka [18]

Answer:

Option (D) is correct.

No, They will not intersect.

Step-by-step explanation:

Given Points for g(x), (1,2) , (2,4) and (3,6)

Since g is a linear function then it must be in form of g(x) = mx +c where, c is constant and m is slope.

We first calculate the slope using,

$m=\frac{y_2-y_1}{x_2-x_1}$

Here, consider $x_1=1 , x_2=2 , y_1=2, y_2=4$

Substitute above, we get slope as,

$m=\frac{4-2}{2-1}=2$

Thus, equation becomes g(x) = 2x + c ...(1)

For c plug the values in (1),

(1,2) ⇒ 2= 2 + c ⇒ c = 0

Thus, Equation of G(x) is 2x.

When we plot it it do not intersect f(x).

Thus, Option (D) is correct.

No, They will not intersect.

3 0

3 years ago

Read 2 more answers

Six more than the product of a number and 5 is equal to 9.

padilas [110]

Answer:

5 = 9 + 6

Step-by-step explanation:

5 0

2 years ago

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

aliya0001 [1]

The Lagrangian

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^4+y^4+z^4-13)$

has critical points where the first derivatives vanish:

$L_x=2x+4\lambda x^3=2x(1+2\lambda x^2)=0\implies x=0\text{ or }x^2=-\dfrac1{2\lambda}$

$L_y=2y+4\lambda y^3=2y(1+2\lambda y^2)=0\implies y=0\text{ or }y^2=-\dfrac1{2\lambda}$

$L_z=2z+4\lambda z^3=2z(1+2\lambda z^2)=0\implies z=0\text{ or }z^2=-\dfrac1{2\lambda}$

$L_\lambda=x^4+y^4+z^4-13=0$

We can't have $x=y=z=0$ , since that contradicts the last condition.

(0 critical points)

If two of them are zero, then the remaining variable has two possible values of $\pm\sqrt[4]{13}$ . For example, if $y=z=0$ , then $x^4=13\implies x=\pm\sqrt[4]{13}$ .

(6 critical points; 2 for each non-zero variable)

If only one of them is zero, then the squares of the remaining variables are equal and we would find $\lambda=-\frac1{\sqrt{26}}$ (taking the negative root because $x^2,y^2,z^2$ must be non-negative), and we can immediately find the critical points from there. For example, if $z=0$ , then $x^4+y^4=13$ . If both $x,y$ are non-zero, then $x^2=y^2=-\frac1{2\lambda}$ , and

$xL_x+yL_y=2(x^2+y^2)+52\lambda=-\dfrac2\lambda+52\lambda=0\implies\lambda=\pm\dfrac1{\sqrt{26}}$

$\implies x^2=\sqrt{\dfrac{13}2}\implies x=\pm\sqrt[4]{\dfrac{13}2}$

and for either choice of $x$ , we can independently choose from $y=\pm\sqrt[4]{\frac{13}2}$ .

(12 critical points; 3 ways of picking one variable to be zero, and 4 choices of sign for the remaining two variables)

If none of the variables are zero, then $x^2=y^2=z^2=-\frac1{2\lambda}$ . We have

$xL_x+yL_y+zL_z=2(x^2+y^2+z^2)+52\lambda=-\dfrac3\lambda+52\lambda=0\implies\lambda=\pm\dfrac{\sqrt{39}}{26}$

$\implies x^2=\sqrt{\dfrac{13}3}\implies x=\pm\sqrt[4]{\dfrac{13}3}$

and similary $y,z$ have the same solutions whose signs can be picked independently of one another.

(8 critical points)

Now evaluate $f$ at each critical point; you should end up with a maximum value of $\sqrt{39}$ and a minimum value of $\sqrt{13}$ (both occurring at various critical points).