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VLD [36.1K]
3 years ago
5

100/h , for h = -25 i need in a sentence

Mathematics
1 answer:
jeka57 [31]3 years ago
8 0

Answer:

-4

Step-by-step explanation:

if h equals -25 the answer is -4

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What is the area of a quadrilateral with vertices at (2,-4), (4,-4), (2,1), and (4, 1).
Rudik [331]
It would be 4,1 squared units so it would be D.
7 0
4 years ago
How many single-scoop cones could be made from a choice of 2 types of cones, 3 flavors of ice cream, and optional toppings of pe
irina1246 [14]

Answer:

Number of different cones made = 18

Step-by-step explanation:

Given:

Types of cone = 2

Number of flavors = 3

Number of toppings = 2

Find:

Number of different cones made

Computation:

Number of different toppings = 1(peanuts) + 1(sprinkles) + 1(both) = 3

Number of different cones made = Types of cone × Number of flavors × Number of different toppings

Number of different cones made = 2 × 3 × 3

Number of different cones made = 18

8 0
4 years ago
Find the area of this composite figure.
Allisa [31]

Answer: ≈ 551.71 ft²

All work is shown in the picture.

5 0
3 years ago
Read 2 more answers
Identify the class​ width, class​ midpoints, and class boundaries for the given frequency distribution.
Crank

Answer:

a. Class width=4

b.

Class midpoints

46.5

50.5

54.5

58.5

62.5

66.5

70.5

c.

Class boundaries

44.5-48.5

48.5-52.5

52.5-56.5

57.5-60.5

60.5-64.5

64.5-68.5

68.5-72.5

Step-by-step explanation:

There are total 7 classes in the given frequency distribution. By arranging the frequency distribution into the refine form we get,

Class

Interval frequency

45-48 1

49-52 3

53-56 5

57-60 11

61-64 7

65-68 7

69-72 1

a)

Class width is calculated by taking difference of consecutive two upper class limits or two lower class limits.

Class width=49-45=4

b)

The midpoints of each class is calculated by taking average of upper class limit and lower class limit for each class.

M=\frac{lower class limit+upper class limit}{2}

Class

Interval Midpoints

45-48 \frac{45+48}{2}=46.5

49-52 \frac{49+52}{2}=50.5

53-56 \frac{53+56}{2}=54.5

57-60 \frac{57+60}{2}=58.5

61-64 \frac{61+64}{2}=62.5

65-68 \frac{65+68}{2}=66.5

69-72 \frac{69+72}{2}=70.5

c)

Class boundaries are calculated by subtracting 0.5 from the lower class limit and adding 0.5 to the upper class interval.

Class

Interval Class boundary

45-48 44.5-48.5

49-52 48.5-52.5

53-56 52.5-56.5

57-60 56.5-60.5

61-64 60.5-64.5

65-68 64.5-68.5

69-72 68.5-72.5

7 0
3 years ago
6. A tank contains 100-gallon of pure water. At time t = 0, a solution containing 2 lb of salt per gallon flows into the tank at
SOVA2 [1]

Answer:

The answer is below

Step-by-step explanation:

Let Q represent the amount of salt in the tank at time t.

\frac{dQ}{dt}= flow\ in - flow\ out\\ \\flow\ in=3\ gal/min*2\ lb/gal=6\ lb/min\\\\net\ gain\ in\ tank\ volume=3-4=-1, henceflow\ out= \frac{4Q}{100-t} \\\\\frac{dQ}{dt}= 6-\frac{4Q}{100-t} \\\\\frac{dQ}{dt}+ \frac{4Q}{100-t}=6\\\\The \ integrating\ factor\ is:\\\\IF=e^{\int\limits {\frac{4}{100-t} } \, dt }=e^{-4\int\limits {\frac{-1}{100-t}}=e^{-4ln(100-t)}=(100-t)^{-4}}\\\\Multiplying\ through  \ by\ IF: \\\\(100-t)^{-4}\frac{dQ}{dt}+ (100-t)^{-4}\frac{4Q}{100-t}=6(100-t)^{-4}\\\\

Integrating:\\\\A(100-t)^{-4}=-2(100-t)^{-3}+c\\\\A=-2(100-t)+\frac{c}{(100-t)^{-4}} \\\\at, t=0,A=0\\\\0=-2(100-0)+\frac{c}{(100-0)^{-4}}\\\\c=0.02\\\\A=-2(100-t)+\frac{0.02}{(100-t)^{-4}}

7 0
3 years ago
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