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3 years ago

Solve! Image attached! Geometry, dumb/incorrect answers will get account, and answer reported, please answer all the way through

Mathematics

1 answer:

neonofarm [45]3 years ago

5 0

Answer:

+) m∠1 = m∠2 (vertical angles theorem)

+) m∠2 and ∠3 are supplementary (because a//b)

+)m∠3 = m∠4 (alternate interior angles theorem, c//d)

=> ∠1 and ∠4 are supplementary (substitution property of equality)

Step-by-step explanation:

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A vegetable garden and a surrounding as a shaped like a square that together a 11 ft wide. The path is 2 feet wide. If one bag o

Alexxandr [17]

Answer:

$\displaystyle 4$

Step-by-step explanation:

we are given that A vegetable garden and a surrounding as a shaped like a square that together a 11 ft wide. The path is 2 feet wide.since together the width of Vegetable garden and path is 11 ft, the width of the vegetables garden will be the difference between the total width and the width of path Thus,

$\displaystyle \rm W _{ garden} = 11 - 2$

simplify substraction:

$\displaystyle \rm W _{ garden} = 9$

recall that, every single side of a square is equal to each other therefore the the area of the garden will be

$\displaystyle {9}^{2}$

simplify square:

$\displaystyle 81$

together the garden and path makes a square of every side length 11 ft saying that the area will be:

$\displaystyle {11}^{2}$

simplify square:

$\displaystyle 121$

the area of path will be the difference between the total area and the garden area therefore,

$\displaystyle 121 - 81$

simplify addition:

$\displaystyle 40$

to figure out how many bags are needed to cover the path. we just need to divide the area of the path by the area of a bag of gravel and that yields:

$\displaystyle \frac{40}{10}$

simplify division:

$\displaystyle \boxed{\rm4}$

hence,

4 bags are needed to cover the path.

7 0

3 years ago

Find the circumference of this circle. 47cm 3.14 30cm 15cm

goldfiish [28.3K]

Answer:

47 cm im nnot sure

Step-by-step explanation:

8 0

3 years ago

Find the two square roots of each complex number by creating and solving polynomial equations.

son4ous [18]

Answer:

1) w₁=4 - i w₂= -4 + i

2) w₁= 3 - i w₂= -3 + i

3) w₁= 1 + 2i w₂= - 1 - 2i

4) w₁= 2- 3i w₂= -2 + 3i

5) w₁= 5 - 2i w₂= -5 + 2i

6) w₁= 5 - 3i w₂= -5 + 3i

Step-by-step explanation:

The root of a complex number is given by:

$\sqrt[n]{z}=\sqrt[n]{r}(Cos(\frac{\theta+2k\pi}{n}) + i Sin(\frac{\theta+2k\pi}{n}))$

where:

r: is the module of the complex number

θ: is the angle of the complex number to the positive axis x

n: index of the root

1) z = 15 − 8i ⇒ r=17 θ= -0.4899 rad

w₁= $\sqrt{17}(Cos(\frac{-0.4899}{2}) + i Sin(\frac{-0.4899}{2}))$ =4-i

w₂= $\sqrt{17}(Cos(\frac{-0.4899+2\pi}{2}) + i Sin(\frac{-0.4899+2\pi}{2}))$ =-1+i

2) z = 8 − 6i ⇒ r=10 θ= -0.6435 rad

w₁= $\sqrt{10}(Cos(\frac{ -0.6435}{2}) + i Sin(\frac{ -0.6435}{2}))$ = 3 - i

w₂= $\sqrt{10}(Cos(\frac{ -0.6435+2\pi}{2}) + i Sin(\frac{ -0.6435+2\pi}{2}))$ = -3 + i

3) z = −3 + 4i ⇒ r=5 θ= -0.9316 rad

w₁= $\sqrt{5}(Cos(\frac{-0.9316}{2}) + i Sin(\frac{-0.9316}{2}))$ = 1 + 2i

w₂= $\sqrt{5}(Cos(\frac{-0.9316+2\pi}{2}) + i Sin(\frac{-0.9316+2\pi}{2}))$ = -1 - 2i

4) z = −5 − 12i ⇒ r=13 θ= 0.4426 rad

w₁= $\sqrt{13}(Cos(\frac{0.4426}{2}) + i Sin(\frac{0.4426}{2}))$ = 2- 3i

w₂= $\sqrt{13}(Cos(\frac{0.4426+2\pi}{2}) + i Sin(\frac{0.4426+2\pi}{2}))$ = -2 + 3i

5) z = 21 − 20i ⇒ r=29 θ= -0.8098 rad

w₁= $\sqrt{29}(Cos(\frac{-0.8098}{2}) + i Sin(\frac{-0.8098}{2}))$ = 5 - 2i

w₂= $\sqrt{29}(Cos(\frac{-0.8098+2\pi}{2}) + i Sin(\frac{-0.8098+2\pi}{2}))$ = -5 + 2i

6) z = 16 − 30i ⇒ r=34 θ= -1.0808 rad

w₁= $\sqrt{34}(Cos(\frac{-1.0808}{2}) + i Sin(\frac{-1.0808}{2}))$ = 5 - 3i

w₂= $\sqrt{34}(Cos(\frac{-1.0808+2\pi}{2}) + i Sin(\frac{-1.0808+2\pi}{2}))$ = -5 + 3i

6 0

3 years ago

Which equation represents the relationship shown in the graph below?

fomenos

I believe the answer is B
:)

3 0

4 years ago

A^3+5-6(4) HELP FAST I’M IN CLASS

zheka24 [161]

A^3-19 should be the answer, good luck

7 0

3 years ago