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2 years ago

15

Y, if m=3, x=6, and b=7

2 answers:

gogolik [260]2 years ago

8 0

Answer:

y=mx+b

y=(3×6)+7

y=18+7

y=25

likoan [24]2 years ago

7 0

Answer:

Step-by-step explanation:

y=mx +b so

y=3(6) + 7

y=18+7

y=25

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Find the least common multiple of 10x4 and 6w3 .

jarptica [38.1K]

10x⁴ and 6w³

1) We see no the same letters in both expressions, so we will need to include all letters in the final answer. x⁴w³

2) number part
10 = 2*5
6=3*2
We need 2*3*5=30 (we need 2 only one times, because we have 2 only one time in each of the numbers).
3)The least common multiple
2*3*5x⁴w³ = 30x⁴w³
Answer is 30x⁴w³.

8 0

2 years ago

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

2 years ago

In which quadrant does the point (-4,15) lie

Ivanshal [37]

Quadrant one

Hope this helped

7 0

2 years ago

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artcher [175]

Answer:

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Step-by-step explanation:

I believe that is <KLJ and <MLJ

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2 years ago

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user100 [1]

Answer:

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Step-by-step explanation:

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2 years ago

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