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2 years ago

7

Dont answer this PLEAS

1 answer:

pickupchik [31]2 years ago

5 0

Answer:

why? also the answer is 1.6y+y-4/15y+7/6y=21/3

We move all terms to the left:

1.6y+y-4/15y+7/6y-(21/3)=0

Domain of the equation: 15y!=0

y!=0/15

y!=0

y∈R

Domain of the equation: 6y!=0

y!=0/6

y!=0

y∈R

We add all the numbers together, and all the variables

1.6y+y-4/15y+7/6y-7=0

We add all the numbers together, and all the variables

2.6y-4/15y+7/6y-7=0

We calculate fractions

2.6y+(-24y)/90y^2+105y/90y^2-7=0

We multiply all the terms by the denominator

(2.6y)*90y^2+(-24y)+105y-7*90y^2=0

We add all the numbers together, and all the variables

(+2.6y)*90y^2+(-24y)+105y-7*90y^2=0

We add all the numbers together, and all the variables

105y+(+2.6y)*90y^2+(-24y)-7*90y^2=0

Wy multiply elements

-630y^2+105y+(+2.6y)*90y^2+(-24y)=0

We get rid of parentheses

-630y^2+105y+(+2.6y)*90y^2-24y=0

We add all the numbers together, and all the variables

-630y^2+81y+(+2.6y)*90y^2=0

Step-by-step explanation:

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Select the correct answer choice below: *<br> Captionless Image

sleet_krkn [62]

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Step-by-step explanation:

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3 years ago

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Use the equation a = IaIâ

german

Answer:

a) $\:\:=\sqrt{14}\cdot \frac{\:\:}{\sqrt{14} }$

b) $\:\:=\sqrt{29} \cdot \frac{\:\:}{\sqrt{29} }$

c) $\:\:=7\cdot \frac{\:\:}{7}$

Step-by-step explanation:

a) Let <u>a</u>=<2,1,-3>

The magnitude of <u>a</u> is $|a|=\sqrt{2^2+1^2+(-3)^2}$

$|a|=\sqrt{4+1+9}=\sqrt{14}$

The unit vector in the direction of a is

$\hat{a}=\frac{\:\:}{\sqrt{14} }$

Using the relation $a=|a|\hat{a}$ , we have

$\:\:=\sqrt{14}\cdot \frac{\:\:}{\sqrt{14} }$

b) Let a=2i - 3j + 4k

$|a|=\sqrt{2^2+(-3)^2+4^2}$

$|a|=\sqrt{4+9+16}=\sqrt{29}$

$\hat{a}=\frac{\:\:}{\sqrt{29} }$

Using the relation $a=|a|\hat{a}$ , we have

$\:\:=\sqrt{29} \cdot \frac{\:\:}{\sqrt{29} }$

c) Let us first find the sum of <1, 2, -3> and <2, 4, 1> to get:

<1+2, 2+4, -3+1>=<3, 6, -2>

Let a=<3, 6, -2>

The magnitude is

$|a|=\sqrt{3^2+6^2+(-2)^2}$

$|a|=\sqrt{9+36+4}=\sqrt{49}=7$

The unit vector in the direction of <u>a</u> is

$\hat{a}=\frac{\:\:}{7}$

Using the relation $a=|a|\hat{a}$ , we have

$\:\:=7\cdot \frac{\:\:}{7}$

5 0

3 years ago