All categories

2 years ago

Check my Work: Factoring for 2y^+9y-18 is (2y-3) (y+6) true or false?

1 answer:

7 0

It is true because when you separate 2y^2 +9y -18, you are supposed to get the factors (2y-3)(y+6), when you multiply this it gives you the first equation, we know this because when we multiply the first term it gives us exactly 2y^2 and if we multiply the last ones it gives you -18, now here is the tricky part, when we multiply the middle part we get 12y and -3y, when we add these up we get 9y, now is that the same middle as in the equation? Yes… then our factors are correct making this true

Hope this helps

You might be interested in

What is 1 2/3 as an inproper fraction?

lesantik [10]

Answer:

5/3

Step-by-step explanation

1 2/3 = 5/3

3/3 + 2/3 = 5/3

6 0

2 years ago

Read 2 more answers

A box of 15 markers cost is $2.40 how much will a box of 40 marker’s cost

butalik [34]

Answer:

$6.40

Step-by-step explanation:

3 0

1 year ago

Read 2 more answers

Which of the following is not one of the 8th roots of unity?

Anika [276]

Answer:

1+i

Step-by-step explanation:

To find the 8th roots of unity, you have to find the trigonometric form of unity.

1. Since $z=1=1+0\cdot i,$ then

$Rez=1,\\ \\Im z=0$

and

$|z|=\sqrt{1^2+0^2}=1,\\ \\\\\cos\varphi =\dfrac{Rez}{|z|}=\dfrac{1}{1}=1,\\ \\\sin\varphi =\dfrac{Imz}{|z|}=\dfrac{0}{1}=0.$

This gives you $\varphi=0.$

Thus,

$z=1\cdot(\cos 0+i\sin 0).$

2. The 8th roots can be calculated using following formula:

$\sqrt[8]{z}=\{\sqrt[8]{|z|} (\cos\dfrac{\varphi+2\pi k}{8}+i\sin \dfrac{\varphi+2\pi k}{8}), k=0,\ 1,\dots,7\}.$

Now

at k=0, $z_0=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 0}{8}+i\sin \dfrac{0+2\pi \cdot 0}{8})=1\cdot (1+0\cdot i)=1;$

at k=1, $z_1=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 1}{8}+i\sin \dfrac{0+2\pi \cdot 1}{8})=1\cdot (\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};$

at k=2, $z_2=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 2}{8}+i\sin \dfrac{0+2\pi \cdot 2}{8})=1\cdot (0+1\cdot i)=i;$

at k=3, $z_3=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 3}{8}+i\sin \dfrac{0+2\pi \cdot 3}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};$

at k=4, $z_4=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 4}{8}+i\sin \dfrac{0+2\pi \cdot 4}{8})=1\cdot (-1+0\cdot i)=-1;$

at k=5, $z_5=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 5}{8}+i\sin \dfrac{0+2\pi \cdot 5}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};$

at k=6, $z_6=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 6}{8}+i\sin \dfrac{0+2\pi \cdot 6}{8})=1\cdot (0-1\cdot i)=-i;$

at k=7, $z_7=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 7}{8}+i\sin \dfrac{0+2\pi \cdot 7}{8})=1\cdot (\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};$

The 8th roots are

$\{1,\ \dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ i, -\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ -1, -\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2},\ -i,\ \dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2}\}.$

Option C is icncorrect.

5 0

2 years ago

A number to the 10th power divided by the same number to the 7th power equals 125. What is the number?

Viefleur [7K]

Answer:

the number is 5

Step-by-step explanation:

x^10 ÷ x^7 = 125

x^3 = 125

x = 5

6 0

2 years ago

What is the solution to this Equation?

Maru [420]

Answer:

D(x=1).

Step-by-step explanation:

2(5x+8)=6x+20

10x+16=6x+20

10-6x+16=6x-6x+20

4x+16=20

4x+16-16=20-16

4x=4

4x/4=4/4

x=1

7 0

1 year ago

Read 2 more answers