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3 years ago

Hey guys i need help plz will give brainlist

Mathematics

1 answer:

nadya68 [22]3 years ago

5 0

Step-by-step explanation:

Effective resistance in the 30Ω and 40Ω parallel circuit = 1 / (1/30 + 1/40) = 120/7 Ω.

20Ω + 10Ω = 30Ω

Effective resistance in the 50Ω and (20Ω + 10Ω) parallel circuit = 1 / (1/50 + 1/30) = 75/4 Ω.

Now we can add the resistances up:

Effective resistance for circuit

= 120/7 Ω + 75/4 Ω = 1005/28 Ω.

Hence I = V / R

= 9.0V / (1005/28 Ω) = 0.25A. (2s.f.)

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A stamp has the Length of 1/2 cm and a width of 1 1/2 is the area of the stamps

avanturin [10]

Answer:

$Area = \frac{3}{4} cm^2$

Step-by-step explanation:

Given

$Length =\frac{1}{2}\ cm$

$Width = 1\frac{1}{2}\ cm$

Required

Determine the Area

Area is calculated as follows

$Area = Length * Width$

Substitute values for Length and Width

$Area = \frac{1}{2} * 1\frac{1}{2}$

Convert mixed fraction

$Area = \frac{1}{2} * \frac{3}{2}$

$Area = \frac{3}{4}$

Hence, the area is

$Area = \frac{3}{4} cm^2$

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2 years ago

What is the product of three square root of eight times four square root of three? Simplify your answer.

igor_vitrenko [27]

$The\ product\ of 3 \sqrt{8}*4 \sqrt{3}\ is\ 24 \sqrt{6}.$

First, recognize that all the numbers can be moved around because they are being multiplied together.
Next, rearrange the expression so that the whole-number coefficients are together:

$3\sqrt{8}*4\sqrt{6}$
$3*4*\sqrt{8}*\sqrt{3}$

Now try to simplify the square-roots to find any squares inside that can be reduced and taken out of the square-root:

$3*4*\sqrt{(2*2*2)}*\sqrt{3}$
$3*4*\sqrt{4*2}*\sqrt{3}$

Multiply the square roots together:
$3*4*\sqrt{(4*2*3)}$
$3*4*\sqrt{(4*6)}$

Now take out the square-root of 4 and simplify
$3*4*\sqrt{4}*\sqrt{6}$
$3*4*(2)*\sqrt{6}$
$24\sqrt{6}$

$Thus,\ your\ answer\ is\ 24\sqrt{6}$

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3 years ago

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3 years ago

4 Ayla s is paid $15 20 per hour for a 34 hour week. April has the same weekly pay but works 38

VashaNatasha [74]

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Now to find Apri’s rate divide the weekly pay by hours:

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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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2 years ago