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3 years ago

PLSSSSSSS HELPPPPPppppp

Mathematics

1 answer:

icang [17]3 years ago

5 0

Answer:

20 units

Step-by-step explanation:

The polygon has 4 vertices as indicated by the 4 coordinate points given.

(3, 1) and (8, 1) is a horizontal side of length 8 - 3 = 5

Similarly

(3, 6) and (8, 6) is the opposite horizontal side of length 8 - 3 = 5

Points (3, 1) and (3, 6) is a vertical side of length 6 - 1 = 5

(8, 1) and (8, 6) is the opposite side of length 6 - 1 = 5

The polygon is therefore a square of side 5 units.

Perimeter = 4 × 5 = 20 units

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Need help with this one also please. Due in half hour. Greatly appreciate it

natka813 [3]

Answer:

9 nickels

Step-by-step explanation:

Changing a quarter to a nickel decreases the value of the sum by $0.20.

When all quarters are changed to nickels and vice versa, the total value will change by the number of excess quarters. That number is ...

($5.95 -3.35)/(0.20) = 2.60/0.20 = 13

The remaining change is made up of equal numbers of nickels and quarters. That amount of change is ...

$5.95 -13·0.25 = 2.70

A quarter and nickel together total $0.30, so there must be $2.70/$0.30 = 9 such pairs of coins.

There are 9 nickels and 22 quarters.

_____

If you don't want to reason through the problem, you can write equations. Let n and q represent the numbers of nickels and quarters, respectively. Then you have ...

0.05n + 0.25q = 5.95

0.25n + 0.05q = 3.35

Multiplying the second equation by 5 and subtracting the first gives ...

5(0.25n +0.05q) -(0.05n +0.25q) = 5(3.35) -(5.95)

1.20n = 10.80

10.80/1.20 = n = 9

The number of nickels is 9.

8 0

3 years ago

What is the point called where the perpendicular bisectors of the sides of a triangle intersect

Vikki [24]

Answer: Circumcenter

This is the center of the circle that goes through the three vertices of the triangle. This circle is known as the circumcircle. It is the smallest circle possible in which the circle encompasses the triangle and none of the triangle spills outside the circle.

6 0

3 years ago

Use (a) the midpoint rule and (b) simpson's rule to approximate the below integral. ∫ x^2sin(x) dx with n = 8.

MaRussiya [10]

Answer:

midpoint rule = 5.93295663

simpson's rule = 5.869246855

Step-by-step explanation:

a) midpoint rule

$\int\limits^b_a {(x)} \, dx$ ≈ Δ x (f(x₀+x₁)/2 + f(x₁+x₂)/2 + f(x₂+x₃)/2 +...+ f(x $_{n}$ _₂+x $_{n}$ _₁)/2 +f(x $_{n}$ _₁+x $_{n}$ )/2)

Δx = (b − a) / n

We have that a = 0, b = π, n = 8

Therefore

Δx = (π − 0) / 8 = π/8

Divide the interval [0,π] into n=8 sub-intervals of length Δx = π/8 with the following endpoints:

a=0, π/8, π/4, 3π/8, π/2, 5π/8, 3π/4, 7π/8, π = b

Now, we just evaluate the function at these endpoints:

$f(\frac{x_{0}+x_{1} }{2} ) = f(\frac{0+\frac{\pi}{8} }{2} ) = f(\frac{\pi }{16})=\frac{\pi^{2}sin(\frac{\pi }{16}) }{256} =$ 0.00752134

$f(\frac{x_{1}+x_{2} }{2} ) = f(\frac{\frac{\pi }{8} +\frac{\pi}{4} }{2} ) = f(\frac{3\pi }{16})=\frac{9\pi ^{2} sin(\frac{3\pi }{16}) }{256}$ = 0.19277080

$f(\frac{x_{2}+x_{3} }{2} ) = f(\frac{\frac{\pi }{4} +\frac{3\pi}{8} }{2} ) = f(\frac{5\pi }{16})=\frac{25\pi ^{2} sin(\frac{5\pi }{16}) }{256}$ = 0.80139415

$f(\frac{x_{3}+x_{4} }{2} ) = f(\frac{\frac{3\pi }{8} +\frac{\pi}{2} }{2} ) = f(\frac{7\pi }{16})=\frac{49\pi ^{2} sin(\frac{7\pi }{16}) }{256}$ = 1.85280536

$f(\frac{x_{4}+x_{5} }{2} ) = f(\frac{\frac{\pi }{2} +\frac{5\pi}{8} }{2} ) = f(\frac{9\pi }{16})=\frac{81\pi ^{2} sin(\frac{7\pi }{16}) }{256}$ = 3.062800704

$f(\frac{x_{5}+x_{6} }{2} ) = f(\frac{\frac{5\pi }{8} +\frac{3\pi}{4} }{2} ) = f(\frac{11\pi }{16})=\frac{121\pi ^{2} sin(\frac{5\pi }{16}) }{256}$ = 3.878747709

$f(\frac{x_{6}+x_{7} }{2} ) = f(\frac{\frac{3\pi }{4} +\frac{7\pi}{8} }{2} ) = f(\frac{13\pi }{16})=\frac{169\pi ^{2} sin(\frac{3\pi }{16}) }{256}$ = 3.61980731

$f(\frac{x_{7}+x_{8} }{2} ) = f(\frac{\frac{7\pi }{8} +\pi }{2} ) = f(\frac{15\pi }{16})=\frac{225\pi ^{2} sin(\frac{\pi }{16}) }{256}$ = 1.69230261

Finally, just sum up the above values and multiply by Δx = π/8:

π/8 (0.00752134 +0.19277080+ 0.80139415 + 1.85280536 + 3.062800704 + 3.878747709 + 3.61980731 + 1.69230261) = 5.93295663

b) simpson's rule

$\int\limits^b_a {(x)} \, dx$ ≈ (Δx)/3 (f(x₀) + 4f(x₁) + 2f(x₂) + 4f(x₃) + 2f(x₄) + ... + 2f( $x_{n-2}$ ) + 4f( $x_{n-1}$ ) + f( $x_{n}$ ))

where Δx = (b−a) / n

We have that a = 0, b = π, n = 8

Therefore

Δx = (π−0) / 8 = π/8

Divide the interval [0,π] into n = 8 sub-intervals of length Δx = π/8, with the following endpoints:

a = 0, π/8, π/4, 3π/8, π/2, 5π/8, 3π/4, 7π/8 ,π = b

Now, we just evaluate the function at these endpoints:

f(x₀) = f(a) = f(0) = 0 = 0

$4f(x_{1} ) = 4f(\frac{\pi }{8} )=\frac{\pi^{2}\sqrt{\frac{1}{2}-\frac{\sqrt{2} }{4} } }{16}$ = 0.23605838

$2f(x_{2} ) = 2f(\frac{\pi }{4} )=\frac{\sqrt{2\pi^{2} } }{16}$ = 0.87235802

$4f(x_{3} ) = 4f(\frac{3\pi }{8} )=\frac{9\pi^{2}\sqrt{\frac{\sqrt{2} }{4}-\frac{{1} }{2} } }{16}$ = 5.12905809

$2f(x_{4} ) = 2f(\frac{\pi }{2} )=\frac{\pi ^{2} }{2}$ = 4.93480220

$4f(x_{5} ) = 4f(\frac{5\pi }{8} )=\frac{25\pi^{2}\sqrt{\frac{\sqrt{2} }{4}-\frac{{1} }{2} } }{16}$ = 14.24738359

$2f(x_{6} ) = 2f(\frac{3\pi }{4} )=\frac{9\sqrt{2\pi^{2} } }{16}$ = 7.85122222

$4f(x_{7} ) = 4f(\frac{7\pi }{8} )=\frac{49\pi^{2}\sqrt{\frac{1}{2}-\frac{\sqrt{2} }{4} } }{16}$ = 11.56686065

f(x₈) = f(b) = f(π) = 0 = 0

Finally, just sum up the above values and multiply by Δx/3 = π/24:

π/24 (0 + 0.23605838 + 0.87235802 + 5.12905809 + 4.93480220 + 14.24738359 + 7.85122222 + 11.56686065 = 5.869246855

7 0

3 years ago

Jeannette's starting salary is $25,000 a year. she will get a 4% raise annually. how much money will she have made after 7 years

Dmitrij [34]

Jeannete would make 32,000 that year.

Hope this helped:D

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3 years ago

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