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3 years ago

Can someone help with trigonometry?? I don't know what these points are

Mathematics

1 answer:

Alisiya [41]3 years ago

3 0

The work is shown above.

I am joyous to assist you anytime.

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Solve the equation 3^(2x) -4*3^(x+1) + 27 = 0. use rules of exponents to change 3^(x+1)

Leviafan [203]

Answer:

$\huge\boxed{x=1\ \vee\ x=2}$

Step-by-step explanation:

$3^{2x}-4\cdot3^{x+1}+27=0\\\\\text{use}\ (a^n)^m=a^{nm}\ \text{and}\ a^n\cdot a^m=a^{n+m}\\\\\left(3^x\right)^2-4\cdot3^x\cdot3^1+27=0\\\\\left(3^x\right)^2-12\cdot3^x+27=0\\\\\text{substitute}\ 3^x=t>0\\\\t^2-12t+27=0\\\\t^2-3t-9t+27=0\\\\t(t-3)-9(t-3)=0\\\\(t-3)(t-9)+0\iff t-3=0\ \vee\ t-9=0\\\\t-3=0\qquad\text{add 3 to both sides}\\\boxed{t=3}\\\\t-9=0\qquad\text{add 9 to both sides}\\\boxed{t=9}$

$\text{We return to substitution:}\\\\3^x=t\\\\3^x=3\ \vee\ 3^x=9\\\\3^x=3^1\ \vee\ 3^x=3^2\\\\\boxed{x=1}\ \vee\ \boxed{x=2}$

4 0

3 years ago

Use the given minimum and maximum data entries, and the number of classes, to find the class width, the lower class limits,

Georgia [21]

Answer:

Step-by-step explanation:

Given that minimum is 8 and maximum equals 82

Range = $84-8 = 76$

No of classes =6

Class width = 76/6 ~13

But not given whether variable is discrete or continuous.

If discrete, we have classes as

8-20, 21-33, 34-46, 47-59, 60-72, 73-85

If continuous, we have classes as

8 to <21

21 to <34

and ... ending 73-<86

7 0

3 years ago

Help with number 3 please

frozen [14]

Answer:

A biconditional statement is a combination of a conditional statement Its converse written in the if and only form

8 0

3 years ago

Let v = (v1, v2) be a vector in R2. Show that (v2, −v1) is orthogonal to v, and use this fact to find two unit vectors orthogona

andrey2020 [161]

Answer:

a. v.v' = v₁v₂ - v₁v₂ = 0 b. (20, -21)/29 and (-20,21)/29

Step-by-step explanation:

a. For two vectors a, b to be orthogonal, their dot product is zero. That is a.b = 0.

Given v = (v₁, v₂) = v₁i + v₂j and v' = (v₂, -v₁) = v₂i - v₁j, we need to show that v.v' = 0

So, v.v' = (v₁i + v₂j).(v₂i - v₁j)

= v₁i.v₂i + v₁i.(- v₁j) + v₂j.v₂i + v₂j.(- v₁j)

= v₁v₂i.i - v₁v₁i.j + v₂v₂j.i - v₂v₁j.j

i.i = 1, i.j = 0, j.i = 0 and j.j = 1

So, v.v' = v₁v₂i.i - v₁v₁i.j + v₂v₂j.i - v₂v₁j.j

= v₁v₂ × 1 - v₁v₁ × 0 + v₂v₂ × 0 - v₂v₁ × 1

= v₁v₂ - v₂v₁

= v₁v₂ - v₁v₂ = 0

So, v.v' = 0

b. Now a vector orthogonal to the vector v = (21,20) is v' = (20,-21).

So the first unit vector is thus a = v'/║v'║ = (20, -21)/√[20² + (-21)²] = (20, -21)/√[400 + 441] = (20, -21)/√841 = (20, -21)/29.

A unit vector perpendicular to a and parallel to v is b = (-21, -20)/29. Another unit vector perpendicular to b, parallel to a and perpendicular to v is thus a' = (-20,-(-21))/29 = (-20,21)/29

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3 years ago

Jerry has an insurance policy with a premium of $150 per month. in june, he's in an accident and receives a bill with a

EastWind [94]

Answer:

he will have to pay $650 in the month of june

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2 years ago

Can someone help with trigonometry?? I don't know what these points are​

Can someone help with trigonometry?? I don't know what these points are