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2 years ago

9

Select equivalent or not equivalent for each pair of expressions

1 answer:

marissa [1.9K]2 years ago

3 0

Answer:

1: equivalent

2: equivalent

3: not equivalent

Step-by-step explanation:

for 1 & 2, you just switch the terms, but for 3, the 6z is not negative in the first expression, but it is in the second.

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The point (−3, 1) is on the terminal side of angle Θ, in standard position. What are the values of sine, cosine, and tangent of

inn [45]

The angle is in the second quadrant so only the sine is positive .
Length of the hypotenuse = sqrt (-1^2 + 3^2) = sqrt10
sine = 1/sqrt10 = 0.3162 to 4 dec places
tangent = 1 / -3 = -0.3333 to 4 d p's
cosine = -3/sqrt10 = -0.9487 to 4 d, p's.

8 0

2 years ago

Read 2 more answers

ZanzabumX [31]

Answer:

To create function h, function f was translated 2 units right , translated 4 units up and reflected across the x axis

Step-by-step explanation:

Given

$f(x) = x^3$

$h(x) =-(x + 2)^3 - 4$

Required

Complete chart

First: f(x) was translated right by 2 units

The rule of right translation is $(x,y) \to (x + 2,y)$

So, we have:

$f'(x) = f(x + 2)$

$f'(x) = (x + 2)^3$

Next: f'(x) was translated up by 4 units

The rule of down translation is $(x,y) \to (x,y+4)$

So, we have:

$f"(x) = f'(x) +4$

$f"(x) = (x + 2)^3 +4$

Lastly, f"(x) was reflected across the x-axis;

The rule of this reflection is: $(x,y) \to (x,-y)$

So, we have:

$h(x) = -f"(x)$

$h(x) = -[(x+2)^3 + 4]$

Remove bracket

$h(x) = -(x+2)^3 - 4$

8 0

2 years ago

the cost of a candle varies directly with the volume of a candle. if a candle with a volume of 80cm^3 costs $12, what would be t

a_sh-v [17]

$120 \div 8. = 1.5$

$12 \times 1.5 = 18$

the answer is $18

good luck

7 0

3 years ago

Help !!!<br>See question in image.<br>Please show workings .<br>

ivolga24 [154]

Answer:

see explanation

Step-by-step explanation:

Given f(x) then the derivative f'(x) is

f'(x) = lim(h tends to 0 ) $\frac{f(x+h)-f(x)}{h}$

= lim ( h to 0 ) $\frac{4(x+h)^2-2-(4x^2-2)}{h}$

= lim ( h to 0 ) $\frac{4(x+h)^2-2-4x^2+2}{h}$

= lim( h to 0 ) $\frac{4x^2+8hx+4h^2-4x^2}{h}$

= lim( h to 0 ) $\frac{8hx+4h^2}{h}$

= lim ( h to 0 ) $\frac{4h(2x+h)}{h}$ ← cancel h on numerator/ denominator

= lim ( h to 0 ) 4(2x + h) ← let h go to zero

f'(x) = 8x

5 0

2 years ago

Which expression could be used to solve the problem?

Paladinen [302]

THE ANSWER IS D whoops lol sorry i was in an argument lol

4 0

2 years ago