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zaharov [31]
3 years ago
15

For a science class, Sweet T counts clusters of bacteria growing in two Perri dishes. He records his finding as a linear functio

n of time

Mathematics
2 answers:
GREYUIT [131]3 years ago
8 0
The correct answer is that "Dish A has a greater rate of change."

In Dish A, the values are going up by 2 each day. In Dish B, the values are staying the same.

Therefore, Dish A has a greater rate of change.
Maru [420]3 years ago
5 0

Answer:

Your answer would be

Dish A has a greater rate of change

Your welcome  e d g e n u i t y  peeps!

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10. A taco stand sells tacos for $3.25 each. The stand’s expenses for the day are $210. Define your variables and write an inequ
makvit [3.9K]
X : number of tacos they sell per day. ...... profit = y = 3.25 × x - 210 ......inequality. y > 0. so. x > 210/3.25 = 840/13 = 64. ......so the taco stand must sell at least 65 tacos for making profit.
8 0
3 years ago
Alan used 9 as an estimate for 3 7/10 + 5 4/10. He added and got 9 1/10 for the actual sum. Is his answer reasonable?
Len [333]

Answer:Yes, he is answer is reasonable

Step-by-step explanation:

First, you plus 3 7/10 with 5 4/8.

Then , when you get 8 11/10 you do add mixed number.

Finally, you get the answer that is 9 1/10

3 0
3 years ago
Chu want some points? Answer correctly n ill give em to ya;)<br><br> 4+7
Andrej [43]

Answer:

11

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
An 18 ft. tree casts a 15 ft. long shadow as shown in the following figure. Which of the following is the angle formed by the su
BabaBlast [244]

Answer:

The angle is 50.2°.

Step-by-step explanation:

The tree and its shadow from a right triangle with the tree being the perpendicular and the shadow being the base as shown in the figure attached.

Therefore the angle θ formed by the sun's rays and the ground is:

\theta = tan^-1(\frac{18}{15} )=\boxed{50.2^o}

8 0
3 years ago
PLSS HELP ILL GIVE YOU BRAINLIEST!!
uysha [10]

Answer:

Step-by-step explanation:

Q1)

Use Phythogoras theorem:

(AC)^2=(BC)^2+(AB)^2\\BC^2=AC^2-AB^2\\BC^2=6^2-4^2\\BC^2=36-16\\BC^2=20\\\\Apply\ square\ on\ both\ sides\\\\\sqrt{BC^2}=\sqrt{20}  \\BC=\sqrt{20} \\BC=\sqrt{10(2)} \\BC=\sqrt{5(2)(2)} \\BC=2\sqrt{5}

Q2)

Apply phythogoras theorem:

CD^2=CE^2+DE^2\\CD^2=7^2+9^2\\CD^2=49+81\\CD^2=130\\\\Apply\ square\ root\ on\ both\ sides\\\\\\sqrt{CD^2} = \sqrt{130} \\\\CD=\sqrt{130} \\\\CD=11.4

Q3)

Apply phythogoras theorem again:

BC^2=BD^2+CD^2\\BC^2=7^2+13^2\\BC^2=49+169\\BC^2=218\\\\Apply\ square\ root\ on\ both\ sides\\\\\\sqrt{BC^2} =\sqrt{218}  \\\\BC=\sqrt{218} \\\\BC=14.76

I have an attached an image for Question 2 for better understanding the length of DE in question equals 15 - 6 = 9

4 0
3 years ago
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