Ab" that goes through points (0,17)

nadya68 [22]

Answer:The United States Electoral College is the group of presidential electors required by the ... They choose one elector per district and two electors for the ticket with the ... on the count of both voter and non-voter residents for state apportionment. ... there can be a president elected who does not win the national popular vote.

Step-by-step explanation:

3 0

3 years ago

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Marco borrowed $150 from his brother.He has paid him back 30% so far.How much money does Marco still owe his brother?

alexdok [17]

150*30% = $45
$150-$45 = $105

Marco still owes his brother $105

8 0

3 years ago

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Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

3 years ago

Is the circumference of all circles about 3 times greater than their diameters. and why?

stira [4]

Answer:

<h2>Yes the circumference is about times greater than their diameters</h2>

Step-by-step explanation:

We will approach this problem by doing a check

Say the diameter of a circle is 4 cm

Hence the radius is 3 cm

We know that the circumference is $C=2\pi r$

$C=2*3.142*2= 12.57cm$

The circumference is about 3 time bigger than the diameter

Let us try another case say the diameter is 6 cm

The radius is 3 cm

$C=2*3.142*3= 18.85cm$

8 0

3 years ago

Compute the probability of the event

Lunna [17]

Answer:

Step-by-step explanation:

joint porbability formula

0.23*0.73*0.74 = 0.124246

7 0

3 years ago