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3 years ago

For the function f(x)=2x+1 and g(x)=3x-2, find (f-g)(4)

Mathematics

1 answer:

icang [17]3 years ago

4 0

Answer:

-5

Step-by-step explanation:

Here is how you solve this problem:

First, you set it up like this: 2x+1-3x+2

Second:You are finding f(4) and g(4) so you just plug in 4 for both problems:

2(4)+1-3(4)-2= 9-14= -5

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Suppose there are approximately 220 calories in 3 servings of a particular brand of cereal and 362 calories in 5 servings of the

ELEN [110]

Answer:

.93 calories per serving

Step-by-step explanation:

220 ÷ 3 = 73.3333333 calories per serving

3 ÷ 220 = 0.0136363636

362 ÷ 5 = 72.4

5 ÷ 362 = 0.0138121547

73.3333333 - 72.4 = 0.9333333 = around .93

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2 years ago

Francis is making cookies. The recipe for chocolate chip cookies calls for 2 2/3 cups of flour, the recipe for peanut butter cal

Alja [10]

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3 years ago

Which equation can be used to solve for x in the following diagram? pls help

Cloud [144]

Answer:b

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Lim t^4 - 6 / 2t^2 - 3t + 7

Harman [31]

I think you meant to say

$\displaystyle \lim_{t\to2}\frac{t^4-6}{2t^2-3t+7}$

(as opposed to x approaching 2)

Since both the numerator and denominator are continuous at t = 2, the limit of the ratio is equal to a ratio of limits. In other words, the limit operator distributes over the quotient:

$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)}$

Because these expressions are continuous at t = 2, we can compute the limits by evaluating the limands directly at 2:

$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)} = \frac{2^4-6}{2\cdot2^2-3\cdot2+7} = \boxed{\frac{10}9}$

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3 years ago

Do these ratios form a proportion?

ad-work [718]

Answer:

Step-by-step explanation:

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3 years ago