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4 years ago

Evaluate the expession. x=12, y=8, z=3 4x-yz

Mathematics

2 answers:

Nataly_w [17]4 years ago

8 0

Given:
<span> x=12, y=8, z=3

4x-yz = 4(12) - (8)(3)
</span>4x-yz = 48 - 24
4x-yz = 24

suter [353]4 years ago

7 0

Answer is 24 ..............;)

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Write in standard form: (5×100)+(6×1)+(5×1/10)+(9×1/1000)

PIT_PIT [208]

Answer:

9,556

Step-by-step explanation:

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3 years ago

Let f(x) = -2x + 7 and g(x) = -6x + 3. Find (f o g) and state its domain.

Vladimir79 [104]

$(f\circ g)(x) = -2(-6x+3) + 7 = 12x - 6 + 7 = 12x + 1$

The domain is all real numbers as a composition of two linear functions have no restrictions (all linear functions have domain of R)

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4 years ago

(x + 6)(x + 2) = 60?

Salsk061 [2.6K]

Answer:

x^3=6

Step-by-step explanation:

x(x+2)+6(x+2)=60

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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

Help me find the solution set for 8x-3=2(x-1/2)

daser333 [38]

8x -3 = 2(x-1/2)
⇒ 8x -3= 2x -2*(1/2)
⇒ 8x -3 = 2x -1
⇒ 8x -2x= -1+ 3
⇒ 6x = 2
⇒ x= 2/6
⇒ x= 1/3

The final answer is x= 1/3~

6 0

3 years ago