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3 years ago

The ordered pair below is a solution to which equation? (2, 12) A. y = 6x B. y=1/2x C. y = 2x D. y=1/6x

2 answers:

7 0

We can substitute the values of x and y to see which equation is satisfied.

A. 12 = 6(2)
12 = 12, true

B. 12 = 1/2 (2)
12 ≠ 1, false

C. 12 = 2(2)
12 ≠ 4, false

D. 12 = 1/6 (2)
12 ≠ 1/3, false

The answer is A.

Illusion [34]3 years ago

3 0

Substitute the x for 2 and the y for 12 in each equation. The first choice works out. (12)=6(2), 12=12. This means the answer is true. A is your answer.

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PLEASE HELP !! ILL GIVE BRAINLIEST *EXTRA POINTS*.. IM GIVING 40 POINTS !! DONT SKIP :((.

stira [4]

Answer: It depends on how many times Tyrell wants to reserve the court. Sorry but there's not enough info to solve rn

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3 0

3 years ago

From home, Mary’s work is two thirds along the way to training. Training is 2.5km from work. Mary normally goes to work, then tr

polet [3.4K]

First thing to do is to illustrate the problem, Since it was mentioned that work was along the way to training, the order is shown in the picture. Mary's home and workplace are nearer compared to her training center. It is also mentioned that the distance between work and home, denoted as x, is 2/3 of the total distance from home to training. The total distance is (x + 2.5). Thus,

x = 2/3(x+2.5)
x = 2/3 x + 5/3
1/3 x = 5/3
x = 5 km

Thus, the distance from home to work is 5 km. This means that Mary has to walk this distance twice to return home to get her shoes. Then, she will travel again the total distance of 5+2.5 = 7.5 km to get to her training center. So,

Total distance = 2(5km) + 7.5 km
Total distance = 17.5 km

3 0

2 years ago

A box designer has been charged with the task of determining the surface area of various open boxes (no lid) that can be constru

Viktor [21]

Answer:

1) $S = 2\cdot w\cdot l - 8\cdot x^{2}$ , 2) The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$ , 3) $S = 176\,in^{2}$ , 4) $x \approx 4.528\,in$ , 5) $S = 164.830\,in^{2}$

Step-by-step explanation:

1) The function of the box is:

$S = 2\cdot (w - 2\cdot x)\cdot x + 2\cdot (l-2\cdot x)\cdot x +(w-2\cdot x)\cdot (l-2\cdot x)$

$S = 2\cdot w\cdot x - 4\cdot x^{2} + 2\cdot l\cdot x - 4\cdot x^{2} + w\cdot l -2\cdot (l + w)\cdot x + l\cdot w$

$S = 2\cdot (w+l)\cdot x - 8\cdpt x^{2} + 2\cdot w \cdot l - 2\cdot (l+w)\cdot x$

$S = 2\cdot w\cdot l - 8\cdot x^{2}$

2) The maximum cutout is:

$2\cdot w \cdot l - 8\cdot x^{2} = 0$

$w\cdot l - 4\cdot x^{2} = 0$

$4\cdot x^{2} = w\cdot l$

$x = \frac{\sqrt{w\cdot l}}{2}$

The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$

3) The surface area when a 1'' x 1'' square is cut out is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1\,in)^{2}$

$S = 176\,in^{2}$

4) The size is found by solving the following second-order polynomial:

$20\,in^{2} = 2 \cdot (8\,in)\cdot (11.5\,in)-8\cdot x^{2}$

$20\,in^{2} = 184\,in^{2} - 8\cdot x^{2}$

$8\cdot x^{2} - 164\,in^{2} = 0$

$x \approx 4.528\,in$

5) The equation of the box volume is:

$V = (w-2\cdot x)\cdot (l-2\cdot x) \cdot x$

$V = [w\cdot l -2\cdot (w+l)\cdot x + 4\cdot x^{2}]\cdot x$

$V = w\cdot l \cdot x - 2\cdot (w+l)\cdot x^{2} + 4\cdot x^{3}$

$V = (8\,in)\cdot (11.5\,in)\cdot x - 2\cdot (19.5\,in)\cdot x^{2} + 4\cdot x^{3}$

$V = (92\,in^{2})\cdot x - (39\,in)\cdot x^{2} + 4\cdot x^{3}$

The first derivative of the function is:

$V' = 92\,in^{2} - (78\,in)\cdot x + 12\cdot x^{2}$

The critical points are determined by equalizing the derivative to zero:

$12\cdot x^{2}-(78\,in)\cdot x + 92\,in^{2} = 0$

$x_{1} \approx 4.952\,in$

$x_{2}\approx 1.548\,in$

The second derivative is found afterwards:

$V'' = 24\cdot x - 78\,in$

After evaluating each critical point, it follows that $x_{1}$ is an absolute minimum and $x_{2}$ is an absolute maximum. Hence, the value of the cutoff so that volume is maximized is:

$x \approx 1.548\,in$

The surface area of the box is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1.548\,in)^{2}$

$S = 164.830\,in^{2}$

4 0

2 years ago

Statistics question; please help.

Alex_Xolod [135]

Answer:

Does this sample provide convincing evidence that the machine is working properly?

Yes.

Step-by-step explanation:

Normal distribution test:

$z=\frac{x- \mu }{ \frac{\sigma}{\sqrt{n}} }=\frac{ (x-\mu)\sqrt{n}}{\sigma}$

Where,

$x: \text{ sample mean}$

$\sigma: \text{ standard deviation}$

$n: \text{ sample size determination}$

$\mu: \text{ hypothesized size of the screw}$

$z=\frac{(1.476-1.5)\sqrt{200} }{0.203 }$

$z=\frac{(-0.024)10\sqrt{2} }{0.203 }$

$z \approx -1.672$

Once the significance level was not given, It is usually taken an assumption of a 5% significance level.

Taking the significance level of 5%, which means a confidence level of 95%, we have a z-value of $\pm 1.96$

Therefore, we fail to reject the null. It means that the hypothesis test is not statistically significant: the average length is not different from 1.5!

7 0

2 years ago

Seriously, I don't like when they put "Not enough information". I feel like there is something missing.

Marizza181 [45]

Hey there Gary!

If Angle DBC is 73, you do subtract it from 180 to get ABD.

180 - 73 = 107

Angle ABD = 107
Angle DBC = 73

Angle ABD > DBC
or
Angle DBC < ABD

The line BD/DB is the same length in both angles, proving line AD is greater than CD. This is also justified by the fact that lines AB and CB are congruent.

I hope this helps, let me know if there's something wrong about my answer.

4 0

2 years ago

Read 2 more answers