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Wewaii [24]
3 years ago
12

If Jon has 26 apples and ate four how many does he have

Mathematics
2 answers:
melisa1 [442]3 years ago
6 0
He has 22 apples if he ate four apple
netineya [11]3 years ago
5 0
22 is it’s the basic
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On a map, the scale shown is 1 inch : 3 miles. If an island is 0.5 square inches on the map, what is the actual area of the isla
gtnhenbr [62]

Answer:

4.5

Step-by-step explanation:

First we need to turn the ratio of 1 : 3 into a fraction

1:3 = 1/3

Next we need to square that fraction

1 x 1 = 1

3 x 3 = 9

Then we turn those numbers into a fraction

1/9

We turn the .5 into a fraction

.5/x

And cross multiply both fraction by each other

1/9 x .5/x

1 x X = X

9 x .5 = 4.5

So

4.5 = X

4 0
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
If t varies 1/s and t=2 when s=60 find the relationship between t and s, b, find the value of t when s=90 c, find s when t=2,1/2
suter [353]

Answer:

t = k \frac{1}{s}  \\ 2 = k. \frac{1}{60} \\ k = 120 \\ k \:  \: is \: constant \: of \: variation \\ relationship \: is  \\ t =  \frac{120}{s} . \\ when \: s \:  = 90 \\ t =  \frac{120}{90}  \\ t =  \frac{4}{3}  \\ when \: t = 2 \frac{1}{2}  \\ s =  \frac{120}{2 \frac{1}{2} }  \\ s = 48

4 0
3 years ago
Select all that have a value of 0.  
vladimir1956 [14]

Answer:

the first third and fifth

Step-by-step explanation:


8 0
3 years ago
Read 2 more answers
What is the product? (3a^2b^7)(5a^3b^8)
vitfil [10]
15a^5 b^15 Is The Answer 
4 0
3 years ago
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