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3 years ago

Points A, B, and C are collinear, and B lies between A and C. If AC = 48, AB = 2x+2, and BC = 3x+6, what is BC?

Mathematics

2 answers:

Vesna [10]3 years ago

8 0

Answer:

The value of BC is 30.

Step-by-step explanation:

Given information: A, B, and C are collinear, B lies between A and C, AC = 48, AB = 2x+2, and BC = 3x+6.

If A, B, and C are collinear, B lies between A and C, then by using segment addition property, we get

$AB+BC=AC$

Substitute AC = 48, AB = 2x+2, and BC = 3x+6 in the above equation.

$(2x+2)+(3x+6)=48$

On combining like terms we get

$(2x+3x)+(2+6)=48$

$5x+8=48$

Subtract 8 from both sides.

$5x=48-8$

$5x=40$

Divide 5 from both sides.

$x=8$

The value of x is 8.

We need to find the value of BC.

$BC=3x+6\Rightarrow 3(8)+6=30$

Therefore the value of BC is 30.

aleksandr82 [10.1K]3 years ago

4 0

Answer:

BC=30

Step-by-step explanation:

AC = 48

AB = 2x+2

BC=3x+6

AC= AB + BC

48 = 2x+2+3x+6

48=5x+8

5x=40

x=8

Hence

BC = 3(8)+6

BC=24+6

BC=30

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2 years ago

What number is represented by 8^-3

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3 years ago

Which equation has the solutions x = -3 ± √3i/2 ?

Maurinko [17]

Answer:Answer is option C : [ $x^{2}$ + 3x + 3 ] =0

Note: None of options matches with given question.

instead of "-3" , there should be "- $\frac{3}{2}$ ".

Step-by-step explanation:

Note: None of options matches with given question.

instead of "-3" , there should be " $\frac{3}{2}$ ".

Here, First thing you have to observe the nature of roots.

∴ x = - $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i and x = - $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$

∴ [ x+( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i) ][ x+( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ]=0

∴ [ $x^{2}$ + x( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i)+ x( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i) + ( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i)( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ]=0

∴ [ $x^{2}$ + $\frac{3}{2}$ x + $\frac{\sqrt{3}}{2}$ ix + $\frac{3}{2}$ x - $\frac{\sqrt{3}}{2}$ ix + (3- $\frac{\sqrt{3}}{2}$ i)(3+ $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + ( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i)( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + $\frac{9}{4}$ - ( $\frac{\sqrt{3}}{2}$ i)( $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + $\frac{9}{4}$ - ( $\frac{3}{4}$ ) $i^{2}$ ] =0

∴ [ $x^{2}$ + 3x + $\frac{9}{4}$ + ( $\frac{3}{4}$ ) ] =0

∴ [ $x^{2}$ + 3x + $\frac{12}{4}$ ] =0

∴ [ $x^{2}$ + 3x + 3 ] =0

Thus, Answer is option C : <em>[ $x^{2}$ + 3x + 3 ] =0 </em>

6 0

3 years ago

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jenyasd209 [6]

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3 years ago

9. Mr. McDonald has $500.00 to spend at a bicycle store. All prices listed

Nataliya [291]

Answer:

This will explain it

Step-by-step explanation:

To answer the question, you need to determine the amount Mr. Traeger has left to spend, then find the maximum number of outfits that will cost less than that remaining amount.

Spent so far:

... 273.98 + 3×7.23 +42.36 = 338.03

Remaining available funds:

... 500.00 -338.03 = 161.97

The cycling outfits are about $80 (slightly less), and this amount is about $160 (slightly more), which is 2 × $80.

Mr. Traeger can buy two (2) cycling outfits with the remaining money.

_____

The remaining money is 161.97/78.12 = 2.0733 times the cost of a cycling outfit. We're sure he has no interest in purchasing a fraction of an outfit, so he can afford to buy 2 outfits.

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2 years ago