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2 years ago

Who wants to do an algebra 1 assignment for me? I will email you the pdf

Mathematics

2 answers:

andrew11 [14]2 years ago

4 0

Answer:

Step-by-step explanation:

People arent on brainly to do others work, they are here to help, NOT DO YOUR WORK

Resepect us more please. We stay on here just to help people. We take time out of our days to help people. So please dont make them do your work. They are here to help not be your personal servant.

Have a wonderful day! :)

tia_tia [17]2 years ago

3 0

Answer:

Step-by-step explanation:

I agree with Crankywolf, we are only here to help, not do your whole assignment just because your too lazy and dum to do it yourself

Bye :)

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Which property was used to simplify the expression?

mixer [17]

$(a^n)^m=a^{n\cdot m}\\\\\text{D. Power of a power property}$

6 0

3 years ago

Find the sum of the first 40 even positive integers.

Pachacha [2.7K]

Answer:

1640

Step-by-step explanation:

Even, positive integers

2+4+6+8+10+12+14+16+18+20+22+24+26+28+30+32+34+36+38+40+42+44+46+48+50+52+54+56+58+60+62+64+66+68+70+72+74+76+78+80 = 1640

6 0

1 year ago

4(v+7) = -6(3v-4) +8v

xxTIMURxx [149]

Answer:

v=−27

Step-by-step explanation:

Since v is on the right side of the equation, switch the sides so it is on the left side of the equation.

−6(3v−4)+8v=4(v+7)

Simplify −6(3v−4)+8v

−10v+24=4(v+7)

Simplify 4(v+7).

−10v+24=4v+28

Move all terms containing v to the left side of the equation.

−14v+24=28

Move all terms not containing v to the right side of the equation.

−14v=4

Divide each term by −14 and simplify.

v=−27

The result can be shown in multiple forms.

Exact Form:

v=−27

6 0

2 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

A value of a car is $24,000. You pay it off by making payments of $6200 a year. Write and equation for when the car would be pay

Damm [24]

Answer: 24,000 / 6,200 = x years

Step-by-step explanation:

The car is valued at $24,000.

Yearly payments are $6,200.

The year that it will be paid off can be dtermined by calculating how long it would take $6,200 to get to $24,000.

You can do that by dividing $24,000 by $6,200.

Assume the number of years for the car to be paid off is x:

24,000 / 6,200 = x years

When solved:

= 24,000 / 6,200

= 3.87 years

4 0

2 years ago