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3 years ago

Please help me!!!!!

1 answer:

5 0

ANSWER:

If one number is a multiple of the other, the lowest common multiple ( LCM ) of two numbers is the number of lower value.

For example, if we need to find the LCM of 2 and 16, the answer would have to be the number with the lower value which is 2.

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Thank you and have a lovely day! <3

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Solve the following system of equations algebraically:<br> Y = x^2 + 8x – 9<br> Y = 2x + 7

julia-pushkina [17]

Use simultaneous equations:
2x + 7 = x^2 + 8x - 9
7 = x^2 + 6x - 9
16 = x^2 + 6x
+ or - 4 = 7x
answer is
-4/7 or 4/7

3 0

3 years ago

Consider 8x2 - 48x = -104. Write the equation so that a = 1: x2 + -6 x = -13 Complete the square: x2 – 6x + 9 = –13 + 9 Factor t

ICE Princess25 [194]

Answer:

The roots of the polynomial are;

3 + 2i

and 3-2i

Step-by-step explanation:

Here, we want to solve the given polynomial using the completing the square method

We start by dividing through by 8

This will give;

x^2 - 6x = -13

To complete the square, we simply divide the coefficient of x by 2 and square it

We have this as -6/2 = -3

square it;; = (-3)^2 = 9

Add it to both sides

x^2 - 6x + 9 = -13 + 9

x^2 - 6x + 9 = -4

(x-3)^2 = -4

Find the square root of both sides

x-3 = ±2i

x = 3 + 2i

or x = 3-2i

8 0

3 years ago

Read 2 more answers

Use the rules for multiplication by powers of 10 to calculate 3 x 100

Yakvenalex [24]

300

because 3 x 1 = 3 +00 = 300... make sense?

4 0

3 years ago

18✖️42=(10✖️40)+(10✖️2)+(8✖️2) <br> True or false

tatyana61 [14]

Answer:

Hope i helped you thanks

3 0

3 years ago

Read 2 more answers

Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

sp2606 [1]

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

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

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.

5 0

3 years ago