Find the range of the graph

dybincka [34]

-5 This is the answer with the work
23
-28
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-5

7 0

3 years ago

9. Mark is filling up crates with watermelons for the food drive. He has large watermelons that weigh 10 pounds each and smaller

abruzzese [7]

9. 10x + 5y is the equation for larger watermelons + smaller watermelons (you did not provide information about the medium watermelons, so we must assume given the options you miswrote one of them). We can hold NO MORE than 500 pounds, so 10x + 5y must be smaller than 500. The best answer is C assuming that "5y" represents your "Medium Watermelons"- because smaller watermelons are stated to be 5 and medium ones should therefore be between 5 and 10, the option provided by A wouldn't make since because 3 pounded watermelons are not "medium" in comparison to the "small ones" that are heavier/bigger. Your best option is C, 10x + 5y < 500, the exact answer technically would be 10x + 5y <= 500.

4 0

3 years ago

Aidan is participating in an 8 mile walk for his favorite charity. The organizers used a coordinate grid to plot the course. The

Travka [436]

Answer:

Aidan is 2 miles far from the ending point when he reaches the water station.

Step-by-step explanation:

The locations of the starting point, water station and ending point are (3, 1), (3, 7) and (3, 9), all expressed in miles. First we determine the distances between starting and ending points and between starting point and water station by the Pythagorean Theorem:

From starting point to ending point:

$D = \sqrt{(3-3)^{2}+(9-1)^{2}}$ (Eq. 1)

$D = 8\,mi$

From starting point to water station:

$d = \sqrt{(3-3)^{2}+(7-1)^{2}}$ (Eq. 2)

$d = 6\,mi$

The distance between the water station and the ending point is:

$s = D-d$ (Eq. 3)

$s = 8\,mi-6\,mi$

$s = 2\,mi$

Hence, Aidan is 2 miles far from the ending point when he reaches the water station.

3 0

3 years ago

Need help<br> Question in the screenshot below<br> NO spam<br> Please show your work

Alja [10]

<h3>Answer: x = 6</h3>

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Work Shown:

$\log_{4}(x+10)+\log_{4}(x-2)=\log_{4}(64)\\\\\log_{4}\left((x+10)(x-2)\right)=\log_{4}(64)\\\\(x+10)(x-2)=64\\\\x^2-2x+10x-20=64\\\\x^2-2x+10x-20-64=0\\\\$