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3 years ago

Find the value f(x)=6xf(2)=[?]

1 answer:

5 0

If f(x) = 6x , then f(2) = 6(2) or f(2) = 12... simply substitute the argument in for the variable in your expression on the right.

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42 points!

kupik [55]

Answer:

Step-by-step explanation:

30x20=600

5 0

3 years ago

Read 2 more answers

The coordinate of centroid of a triangle whose vertices are (1,3,-2), (4,5,0), (6,3,9) is = ……………….

Dominik [7]

Answer:

$C = (\frac{11}{3},\frac{11}{3},\frac{7}{3})$

Step-by-step explanation:

Given

$(x_1,y_1,z_1) = (1,3,-2)$

$(x_2,y_2,z_2) = (4,5,0)$

$(x_3,y_3,z_3) = (6,3,9)$

Required

Determine the coordinates of the centroid

Represent the coordinates with C.

C is calculated as follows:

$C = (\frac{1}{3}(x_1+x_2+x_3),\frac{1}{3}(y_1+y_2+y_3),\frac{1}{3}(z_1+z_2+z_3}))$

Substitute values of x and y in the given equation

$C = (\frac{1}{3}(1+4+6),\frac{1}{3}(3+5+3),\frac{1}{3}(-2+0+9}))$

$C = (\frac{1}{3}(11),\frac{1}{3}(11),\frac{1}{3}(7}))$

$C = (\frac{11}{3},\frac{11}{3},\frac{7}{3})$

<em>The above is the coordinate of the centroid</em>

8 0

3 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

3 years ago

Alfonso and Lucas are 238 miles apart, traveling towards each other. If Lucas travels 21 mph and Alfonso travels 13 mph how long

Alex

Lucas equation for his location is x(t) = 13t.

Alfonso's equation is x(t) = 238 - 21t

If you equate these, you get:

13t = 238 - 21t =>

34t = 238

t = 7

So after seven hours they meet.

7 0

3 years ago

How do I calculate 5 percent of 60 using fractions

irinina [24]

60*5%=60*5/100=300/100=3
So the answer is 3

5 0

3 years ago

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