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3 years ago

I need help with this

Mathematics

2 answers:

san4es73 [151]3 years ago

7 0

1. -9
2. -2
3. 3
4. 2

Positive + positive = positive
Negative + negative = negative
Negative + positive = subtract and take sign of bigger number

netineya [11]3 years ago

4 0

Answer:

1. -9

2. -2

3. -3

4. 2 (positive)

Step-by-step explanation:

positive + negative= positive

negative + positive =negative

positive +positive =positive

negative +negative =negative

hope you like it ( ' - ' )

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2 years ago

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3 years ago

Find the value of 9x + 2 given that -5x-3=2

Sloan [31]

$-5x-3 = 2\\-5x=5\\x = -1$

Now plug the x value into 9x + 2

$9(-1)+2) = -7$

Hope that helps!

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2 years ago

At the Science Museum, there is an exhibition with 150 displays about energy.

tatuchka [14]

<span>1/5

This is a matter of elimination. So let's figure out how many of each display there is. 30% of 150 = 0.30 * 150 = 45 displays about electricity. 2/5 * 150 = 0.40 * 150 = 60 displays about renewable energy 1/10 * 150 = 150/10 = 15 displays about gas. The rest of the displays are about nuclear power. So 150 - 45 - 60 - 15 = 30 displays about nuclear power. Since we want a fraction, we get 30/150 But 30/150 isn't in it's simplest form. The largest number that will evenly divide both 30 and 150 is 30. So divide both the numerator and denominator by 30 (30/30)/(150/30) = 1/5 So 1/5 of the displays are about nuclear power.</span>

8 0

3 years ago

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