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Mazyrski [523]
3 years ago
10

For homework a student had to complete 15 problems total. If she finished 6 problems in class, what is the ratio of problems she

still needs to complete to problems that she’s already finished
Mathematics
2 answers:
Korvikt [17]3 years ago
7 0

Answer:

Step-by-step explanation:

so she has 15 in total she completed 6 questions already so she has completed 6:15 for ratio of how many she did and 15-6=9 so she has 9:15 questions left to complete.

6:15 = ratio for questions she has already completed

9:15 = ratio for how many questions she has left

9:15 and 6:15 can be simplified to 3:5

HOPE THIS HELPED (:

nexus9112 [7]3 years ago
6 0

Well if the student already finished 6 problems out of 15, there are 15-6 = 9 homework questions she still has to finish.

Still needs to complete: 9.

Already finished: 6

9 : 6.

Simplified: 3 : 2

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OlgaM077 [116]
First:The area of a circle is πr squared. Not r2
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6 0
3 years ago
3x + y = 23<br> 3x - 2y = 8
sergey [27]

Answer:

5

Step-by-step explanation:

difference in y is 3

difference in answer is 15

15 divided by 3 is 5

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What is the equation of a line that passes through the point (2, 7) and is perpendicular to the line whose equation is y=x4+5 ?
Valentin [98]
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3 years ago
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
3 years ago
ILL MARK BRAINIEST IF YOU DO THIS CORRECTLY!!!
ehidna [41]

Answer:

76 percent done.

Step-by-step explanation:

Reasoning for this if its 76/100 you know that it will be 76. Or im Just got confused.

7 0
2 years ago
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