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3 years ago

QF Q6.) Find the function for b.

Mathematics

2 answers:

Kay [80]3 years ago

8 0

(g₀f)(x)=g(f(x))
=g(2x-4)
=4(2x-4)^2-3
=4(4x^2-16x+16)-3
=16x^2-64x+61

Dimas [21]3 years ago

7 0

The function notation f(x)=2x-4 means replace the x on the right-hand side with the value in parentheses on the left.
For example, f(y)=2y-4; f(4)=2(4)-4=4; f(1)=2(1)-4=-2 and so on.
The same principle applies to g(x), for example
g(2)=4(2^2)-3=13,
g(x+1)=4(x+1)^2-3=4(x^2+2x+1)-3=4x^2+8x+4-3=4x^2+8x+1
and so on.

(g of f)(x) is another way to write g(f(x)), therefore
g(f(x))
=g(2x-4) [because f(x)=2x-4
=4(2x-4)^2-3
=4(4x^2-16x+16)-3
=16x^2-64x+64-3
=16x^2-64x+61

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salantis [7]

Answer:

see explanation

Step-by-step explanation:

Using the sum/ difference → product formula

cos x - cos y = - 2sin( $\frac{x+y}{2}$ )sin ( $\frac{x-y}{2}$ )

sin x - sin y = 2cos ( $\frac{x+y}{2}$ )sin ( $\frac{x-y}{2}$ )

Given

(cosA - cosB)² + (sinA - sinB )²

= [ - 2sin( $\frac{A+B}{2}$ )sin( $\frac{A-B}{2}$ ) ]² + [ 2cos( $\frac{A+B}{2}$ )sin( $\frac{A-B}{2}$ ) ]²

= 4sin² ( $\frac{A+B}{2}$ )sin² ( $\frac{A-B}{2}$ ) + 4cos² ( $\frac{A+B}{2}$ )sin² ( $\frac{A-B}{2}$ )

= 4sin² ( $\frac{A-B}{2}$ )[ sin² ( $\frac{A+B}{2}$ ) + cos² ( $\frac{A+B}{2}$ ) ← sin²x + cos²x = 1

= 4sin² ( $\frac{A-B}{2}$ ) × 1

= 4sin² ( $\frac{A-B}{2}$ ) = right side ⇒ proven

4 0

3 years ago

For triangle ABC, find the measure of AB given me∠ A = 55 degrees, m∠b= 44 degrees, and b=6. . A. 45.22. B.96.68. C.88.19. .

jekas [21]

There exist a similar question where b = 68 instead of 6. First, determine the measure of the third angle, angle C,
m∠c = 180 - (55° + 44°) = 81°
Let x be the side AB, that which is opposite to angle C. Through the Sine Law,
68 / sin 44° = x / sin 81°
From the equation, the value of x is equal to 96.68. Thus, the answer is letter B.

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3 years ago

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lakkis [162]

Step-by-step explanation:

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1/2 × (30×4) × 39 + (30×30)

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2 years ago

I need help with this question i dont understand it

RUDIKE [14]

The perimeter, by definition, is the outside measure of that figure. MN and LM are the same length and LK and NK are the same length....we just need to find the lengths! Use the distance formula to find the distance between the 2 points:
$\sqrt{( x_{2} - x_{1} ) ^{2} +( y_{2}- y_{1} ) ^{2} }$
For the segment MN, use the coordinates of M as your x1, y1, and use the coordinates of N for x2, y2:
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which simplifies to
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So that is the length of both MN and LM. So far our perimeter is $\sqrt{2} + \sqrt{2}=2 \sqrt{2}$
Now let's use the same formula to find out the length of one of the longer segments:
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So our perimeter is, in the end, $2 \sqrt{2}+2 \sqrt{5}$
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3 years ago

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kvv77 [185]

Answer:

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Step-by-step explanation:

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3 years ago