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3 years ago

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Small boxes contain DVDs, and large boxes contain one gaming machine. Three boxes of gaming machines and a box of DVDs weight 48

pounds. Three boxes of gaming machines and five boxes of DVDs weight 72 pounds. How much does each box weigh?
DVD box weighs 14 pounds and gaming machine box weighs 6 pounds
DVD box weighs 6 pounds and gaming machine box weighs 14 pounds
DVD box weighs 40 pounds and gaming machine box weighs 0 pounds
DVD box weighs 8 pounds and gaming machine box weighs 24 pounds

1 answer:

____ [38]3 years ago

7 0

Answer:

The correct Answer to this question will be D, last option

Step-by-step explanation:

x = DVD boxes

y = Gaming boxes

3x + 1y = 48 pounds

Now lets count:

In the Last option we see that DVD boxes have the weight of 8 pounds per box so 3 times 8 = 24, In the last option we see that 1 gaming box weight's 24 pounds

now we add the pounds together 24 p + 24 p = 48 p

So the Answer to your question will be D ot the last option.

I hope it helped someone.

Can I have please have brainliest so people know what answer to choose)

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Find the 98th term of the arithmetic sequence 8, 24, 40,...

Annette [7]

Answer:

Step-by-step explanation:

AP = 8, 24 , 40....

a = 8 , d =16 , n =98 , An = ?

an= a +(n-1)d

an = 8+97*16

an=8+1552

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3 years ago

Find the value of x (3/5)^x(5/3)^2x=125/27

lesya692 [45]

Given:

The equation is

$\left(\dfrac{3}{5}\right)^x\left(\dfrac{5}{3}\right)^{2x}=\dfrac{125}{27}$

To find:

The value of x.

Solution:

We have,

$\left(\dfrac{3}{5}\right)^x\left(\dfrac{5}{3}\right)^{2x}=\dfrac{125}{27}$

$\left(\dfrac{5}{3}\right)^{-x}\left(\dfrac{5}{3}\right)^{2x}=\dfrac{5\times 5\times 5}{3\times 3\times 3}$ $[\because \left(\dfrac{a}{b}\right)^{-m}=\left(\dfrac{b}{a}\right)^{m}]$

$\left(\dfrac{5}{3}\right)^{-x+2x}=\dfrac{5^3}{3^3}$ $[\because a^ma^n=a^{m+n}]$

$\left(\dfrac{5}{3}\right)^{x}=\left(\dfrac{5}{3}\right)^{3}$

On comparing the exponents, we get

$x=3$

Therefore, the value of x is 3.

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3 years ago

PLEASE ANSWER ASAP

SOVA2 [1]

Answer:

x - 2 ≥ 0

Step-by-step explanation:

The expression under the square root cannot be negative, thus

x - 2 ≥ 0 , that is

x ≥ 2

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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.

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3 years ago

What is the first term in the expansion of the binomial (3x+y)^4 ?

solong [7]

81 $x^{4}$

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3 years ago

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