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Nitella [24]
3 years ago
11

2x(x - 1) + 3x – 3 = 0​

Mathematics
2 answers:
xxTIMURxx [149]3 years ago
4 0

Answer:

x = -3/2 or x = 1

Step-by-step explanation:

lozanna [386]3 years ago
3 0

Answer:

x = -3/2 or x = 1

Step-by-step explanation:

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Number 6:
ivolga24 [154]

The answer is X=5 and Y =-2

View the picture to see step by step solution

8 0
3 years ago
What are 3 ways to solve 12×15 math problems
Georgia [21]

Answer:

1- Do 12 times 5 and then 12 times 10. add the sums together

2-Do 15 times 2 and then 15 times 10. Add the sums together

3-You could split 12 in half to get two 6s. Do 15 times 6 and multiply the product by 2 to get your answer

Step-by-step explanation:

7 0
4 years ago
Read 2 more answers
Calculate the volume of a cylinder that is 5 m long and<br> 40 cm wide.
Luden [163]

Answer:

2.51 cubic meters

Step-by-step explanation:

plug into formula calculator.

6 0
3 years ago
This week, Jim practiced the piano 1 1/8 hours on Monday and 2 3/7 hours on Tuesday. How many hours did he practice this week? H
Ainat [17]

Let's make these two fractions have a common denominator and an improper fraction:

1 1/8 = 9/8 =  63/56

2 3/7 = 17/7 = 136/56

63/56 + 136/56 = 199/56 = 3 31/56

So in total, he practiced 3 31/56 hours this week.

He practiced 1 17/56 hours longer on Tuesday than Monday.

3 0
3 years ago
Read 2 more answers
(1-i)^2 find 4 th root​
sergiy2304 [10]

By de Moivre's theorem,

1 - i = \sqrt2\,e^{-i\pi/4} \implies (1-i)^2 = 2\,e^{-i\pi/2}

\implies \sqrt[4]{(1 - i)^2} = \sqrt[4]{2}\,e^{i(2\pi k-\pi/2)/4} = \sqrt[4]{2}\,e^{i(4k-1)\pi/8}

where k\in\{0,1,2,3\}. The fourth roots of (1-i)^2 are then

k = 0 \implies \sqrt[4]{2}\,e^{-i\pi/8}

k = 1 \implies \sqrt[4]{2}\,e^{i3\pi/8}

k = 2 \implies \sqrt[4]{2}\,e^{i7\pi/8}

k = 3 \implies \sqrt[4]{2}\,e^{i11\pi/8}

or more simply

\boxed{\pm\sqrt[4]{2}\,e^{-i\pi/8} \text{ and } \pm\sqrt[4]{2}\,e^{i3\pi/8}}

We can go on to put these in rectangular form. Recall

\cos^2(x) = \dfrac{1 + \cos(2x)}2

\sin^2(x) = \dfrac{1 - \cos(2x)}2

Then

\cos\left(-\dfrac\pi8\right) = \cos\left(\dfrac\pi8\right) = \sqrt{\dfrac{1 + \cos\left(\frac\pi4\right)}2} = \sqrt{\dfrac12 + \dfrac1{2\sqrt2}}

\sin\left(-\dfrac\pi8\right) = -\sin\left(\dfrac\pi8\right) = -\sqrt{\dfrac{1 - \cos\left(\frac\pi4\right)}2} = -\sqrt{\dfrac12 - \dfrac1{2\sqrt2}}

\cos\left(\dfrac{3\pi}8\right) = \sin\left(\dfrac\pi8\right) = \sqrt{\dfrac12 - \dfrac1{2\sqrt2}}

\sin\left(\dfrac{3\pi}8\right) = \cos\left(\dfrac\pi8\right) = \sqrt{\dfrac12 + \dfrac1{2\sqrt2}}

and the roots are equivalently

\boxed{\pm\sqrt[4]{2}\left(\sqrt{\dfrac12 + \dfrac1{2\sqrt2}} - i\sqrt{\dfrac12 - \dfrac1{2\sqrt2}}\right) \text{ and } \pm\sqrt[4]{2}\left(\sqrt{\dfrac12 + \dfrac1{2\sqrt2}} + i \sqrt{\dfrac12 - \dfrac1{2\sqrt2}}\right)}

7 0
2 years ago
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