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lutik1710 [3]
3 years ago
12

3/5+(-7/5)= can someone help me with this please

Mathematics
1 answer:
Tresset [83]3 years ago
4 0

Answer:

-\frac{4}{5}

Step-by-step explanation:

Because both of the fractions have a common denominator they can be subracted without any other work.

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it takes shonda 2/5 of an hour to dry one load of laundry in her dryer. How many hours will it take her to dry 5 loads of laundr
horrorfan [7]

2 hours Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
A function g(x) = -x^2 + 3x - 2. What is g(-4)?
Butoxors [25]
G(-4) = 4^2 + 3(-4) - 2

16 - 12 - 2

4 - 2

answer: 2
7 0
2 years ago
A sample of gold has a mass of 579 g. The volume of the sample is 30 cm3. What is the density of the gold sample?
Ugo [173]

The density of the gold sample is 19.3g/cm³

Let the mass of the gold sample be represented by m

m = 579 g

Let the volume of the gold sample be represented by V

V = 30 cm³

Let the density of the gold sample be represented by ρ

The formula for the density is:

Density = \frac{Mass}{Volume}

\rho = \frac{m}{V} \\\rho = \frac{579}{30} \\\rho = 19.3 g/cm^3

The density of the gold sample is 19.3g/cm³

Learn more here: brainly.com/question/17780219

3 0
3 years ago
Find the area of the shaded regions. Give your answer as a completely simplified exact value in terms of π (no approximations).
olga_2 [115]

Answer:

7\pi cm^2

Step-by-step explanation:

The area of a circle is given by A=\pi r^2

For circle B, the radius is 3 cm, it has an area of \pi *3^2=9\pi cm^2

Circle D also has a radius of 1 cm. Its area is \pi *1^2=\pi cm^2

Circle P also has a radius of 1 cm and area \pi*1^2=\pi cm^2

The area of the shaded region is then 9\pi -\pi-\pi=7\pi cm^2

5 0
3 years ago
Lawrence performs a survey to determine the average number of minutes of exercise seventh-grade athletes get in one week. He get
RoseWind [281]

Answer:

1) Is more representative

Step-by-step explanation:

The problem with his selection is that maybe there are few students participating in certain sport and those students maybe do quite more excercise than the rest (or quite less). This will modify the results because the sample he selected is biased. This problem wont be solved by method 3 or 4, because he is still selecting students that may modify heavily the results with a high probability

This problem will also appear if he choose a sample by class. Maybe, in a class there are quite few students, and selecting from class will make those students appear quite more often than, lets say, a 7th grade student selected at random, therefore the selection is biased in this case as well.

If he has a list with all seventh grade students, each student is equallly likely to be selected and as a consequence, the the results wont be biased. Approach 1 is the best one.

5 0
3 years ago
Read 2 more answers
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