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3 years ago

How does the graph of f(x)=4cos(1/2x)-3 differ from the graph of g(x)=4cos(x)-3

Mathematics

1 answer:

almond37 [142]3 years ago

8 0

Answer:

what is this in? please let me kno2

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At the start of a party, a drink dispenser contains 9 1.2 quarts of iced tea. After the party only 7 cups of tea remain. How man

xxMikexx [17]

Answer: 31 cups

Step-by-step explanation:

We should note that 1 quarts = 4 cups

Therefore, 9 1/2 quarts = 9 1/2 × 4 = 19/2 × 4 = 38 cups

Since only 7 cups of tea remained.after the party, the number of cups of iced tea that the people drank during the party will be:

= 38 - 7

= 31 cups of iced tea

4 0

2 years ago

For the data set 7, 10, 10, 14, 14, 17, the mean is 12. What is the mean

anastassius [24]

Answer: The mean absolute devastation is 3

Step-by-step explanation:

4 0

3 years ago

I need to check if it's correct. Can someone tell me how to explain the whole thing

algol13

One plus one u need to subscribe

4 0

2 years ago

16) Four students simplified the expression given below :(6 + 5 × 4 ) ÷ 2 , Anshu's solutions is 25 , Ram's solution is 46, Kish

Juli2301 [7.4K]

Answer:

Step-by-step explanation:

6 0

3 years ago

Is the line through points P(–3, –2) and Q(2, 3) perpendicular to the line through points R(10, –1) and S(15, –6)? Explain.

Alexus [3.1K]

Answer:

B. Yes. Their slopes have product -1.

Step-by-step explanation:

Given:

Line passing through P(-3, -2) and Q(2, 3)

Line passing through R(10, -1) and S(15, -6)

Required:

To determine if both lines are perpendicular.

SOLUTION:

Two lines are considered perpendicular if the product of their slopes equal -1.

To determine if both lines given in the question are perpendicular, first, calculate their slope using: $m = \frac{y_2 - y_1}{x_2 - x_1}$

Slope of line passing through P(-3, -2) and Q(2, 3):

Let,

$P(-3, -2) = (x_1, y_1)$

$Q(2, 3) = (x_2, y_2)$

$m = \frac{3 -(-2)}{2 -(-3)}$

$m = \frac{5}{5} = 1$

Slope of line passing through R(10, -1) and S(15, -6):

Let,

$R(10, -1) = (x_1, y_1)$

$S(15, -6) = (x_2, y_2)$

$m = \frac{-6 -(-1)}{15 - 10}$

$m = \frac{-5}{5} = -1$

The product of their slopes = 1 × -1 = -1

Therefore, the lines are perpendicular.

The answer is: B. "Yes. Their slopes have product -1."

5 0

3 years ago