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Natalija [7]
2 years ago
9

HELPFAST What is the sum of -2/6 and 3/2? 10 points

Mathematics
2 answers:
Svetllana [295]2 years ago
7 0
7/6 i think! hope this helps
Phantasy [73]2 years ago
3 0

Answer:

-2/6+3/2

-2/6+9/6

7/6

I hope this is good enough:

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Which is the function represented by the table?
Makovka662 [10]

Answer:

A

Step-by-step explanation:

We are given a table with the points [ (-1, 4), (0, 5), (1, 6), (2, 7) ]

We need to find the slope with [ y2-y1/x2-x1 ]

5-4/0-(-1)

1/1

1

The point (0, 5) represents the y-intercept.

y = 1x + 5

y = x + 5

Note that f(x) = y

f(x) = x + 5

Best of Luck!

4 0
3 years ago
Read 2 more answers
museum passes cost $5 for adults and $2 for children. one day the museum sold 1820 passes for $6100. how many of each type were
mr Goodwill [35]
1820 divide by 5 =364
1820 divide by 2=910

3 0
3 years ago
Simplify the expression (5÷5) ÷ 5 * (5^2 - 5)
Katyanochek1 [597]

[ Answer ]

\boxed{\bold{4}}

[ Explanation ]

  • Simplify / Solve: (5 ÷ 5) ÷ 5 · (5^{2} - 5)

--------------------------------

  • Rewrite

\frac{\frac{5}{5} }{5}(5^{2} - 5)

  • Apply Rule (\frac{a}{a} = 1): \frac{5}{5} = 1

\frac{1}{5}(5^{2} - 5)

  • Multiply Fractions

\frac{1 \ * \ 5^{2 \ - \ 5} }{5}

  • Simplify

1 · (5^{2} - 5) = 20

  • Simplify

= \frac{20}{5}

  • Divide

20 ÷ 5

  • Answer

= 4

\boxed{\bold{[] \ Eclipsed \ []}}

3 0
3 years ago
A spring has a natural length of 7 m. If a 4-N force is required to keep it stretched to a length of 11 m, how much work W is re
bezimeni [28]

Answer:

18 J is the work required to stretch a spring from 7 m to 13 m.

Step-by-step explanation:

The work done is defined to be the product of the force F and the distance d  that the object moves:

W=Fd

If F is measured in newtons and d<em> </em>in meters, then the unit for is a newton-meter, which is called a joule (J).

This definition work as long as the force is constant, but if the force is variable like in this case, we have that the work done is given by

W=\int\limits^b_a {f(x)} \, dx

Hooke’s Law states that the force required to maintain a spring stretched x    units beyond its natural length is proportional to

f(x)=kx

where k is a positive constant (called the spring constant).

To find how much work W is required to stretch it from 7 m to 13 m you must:

Step 1: Find the spring constant

We know that the spring has a natural length of 7 m and a 4 N force is required to keep it stretched to a length of 11 m. So, applying Hooke’s Law

4=k(11-7)\\\\\frac{k\left(11-7\right)}{4}=\frac{4}{4}\\\\k=1

Thus F=x

Step 2: Find the the work done in stretching the spring from 7 m to 13 m.

Recall that the natural length is 7 m, so when we stretch the spring from 7 m to 13 m, we are stretching it by 6 m beyond its natural length.

Work needed to stretch it by 6 m beyond its natural length

W=\int\limits^6_0 {x} \, dx \\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1\\\\\left[\frac{x^{1+1}}{1+1}\right]^6_0\\\\\left[\frac{x^2}{2}\right]^6_0=18

18 J is the work required to stretch a spring from 7 m to 13 m.

5 0
3 years ago
PLEASE HELP ASAP GEOMETRY 20 PTS REAL ANSWERS ONLY
kodGreya [7K]

Answer:

Step-by-step explanation:

Use the Law of Cosines to find the measure of angle A from the lengths of the sides.

A = arccos[(b²+c²-a²)/(2bc)] ≅ 29.9°

B = arccos[(a²+c²-b²)/(2ac)] ≅ 54.8°

C = 180 - A - B = 95.3°

7 0
2 years ago
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