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3 years ago

Which points are on the graph of the function rule f(x) = 10 – 4x?

Mathematics

2 answers:

kondor19780726 [428]3 years ago

8 0

(–2, 18), (0, 10), (2, 2)

FromTheMoon [43]3 years ago

4 0

Answer with Step-by-step explanation:

we are given that:

f(x)=10-4x

We have to determine which points lie on the graph of f(x)

1. (2, –18), (0, –10), (–2, –2)

When x=0

f(x)=10

But here we are given (0,-10)

Hence, this option is incorrect

2. (18, –2), (10, 0), (2, 2)

When x=10

f(x)= -30

but here we are given (10,0)

Hence, this option is incorrect

3. (–18, 2), (–10, 0), (–2, –2)

when x= -10

f(x)= 50

but here we are given (-10,0)

Hence, this option is incorrect

4. (–2, 18), (0, 10), (2, 2)

f(-2)=18,f(0)=10 and f(2)=2

Hence, Correct option is:

(–2, 18), (0, 10), (2, 2)

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(4 x 300) + (4 x 25)

Dominik [7]

Answer:

1,300

Step-by-step explanation:

Order of operations tells us that we should do everything in parentheses first.

Multiplying out (4 x 300) and (4 x 25) gets

1200 + 100

From here, we add the two together to arrive at

1,300

3 0

3 years ago

Read 2 more answers

Find the indicated limit, if it exists.

kondor19780726 [428]

Answer:

d) The limit does not exist

General Formulas and Concepts:

Calculus

Limits

Right-Side Limit: $\displaystyle \lim_{x \to c^+} f(x)$
Left-Side Limit: $\displaystyle \lim_{x \to c^-} f(x)$

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

Limit Property [Addition/Subtraction]: $\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$

Step-by-step explanation:

*Note:

In order for a limit to exist, the right-side and left-side limits must equal each other.

Step 1: Define

Identify

$\displaystyle f(x) = \left\{\begin{array}{ccc}5 - x,\ x < 5\\8,\ x = 5\\x + 3,\ x > 5\end{array}$

Step 2: Find Right-Side Limit

Substitute in function [Limit]: $\displaystyle \lim_{x \to 5^+} 5 - x$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \lim_{x \to 5^+} 5 - x = 5 - 5 = 0$

Step 3: Find Left-Side Limit

Substitute in function [Limit]: $\displaystyle \lim_{x \to 5^-} x + 3$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \lim_{x \to 5^+} x + 3 = 5 + 3 = 8$

∴ Since $\displaystyle \lim_{x \to 5^+} f(x) \neq \lim_{x \to 5^-} f(x)$ , then $\displaystyle \lim_{x \to 5} f(x) = DNE$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

5 0

3 years ago

What is the solution to the system? 1. x-y + 2 z = -7 2. y + z =1 3. x-2 y - 3 z = 0

kvv77 [185]

You'd find this problem easier to understand and do if you'd please list the defining equations vertically and line up variables:

1. x - 1y + 2 z = -7

2. y + 1z = 1

3. x - 2 y - 3 z = 0 Now eliminate the line numbers:

x - 1y + 2 z = -7

1y + 1 z = 1

x - 2 y - 3 z = 0

Let's use the elimination method to eliminate variable z: Seeing that z = 1 - y, we transform the first equation into 1x - 1y + 2(1-y) = -7

and the third into x - 2y - 3(1-y) = 0.

Simplifying 1x - 1y + 2(1-y) = -7

and x - 2y - 3(1-y) = 0,

we get

1x - 2y - 3 + 3y) = 0 and 1x - 1y + 2 - 2y = -7

which in turn simplify to

1x + y = 3 and 1x - 3y = -9

Having eliminated the variable z, we now focus on eliminating x. Mult. the 1st equation by -1, obtaining -1x - 1y = -3. Add this result to 1x - 3y = -9:

0 - 4y = -12, which tells us that y = 3. Subbing 3 for y in 1x + 1y = 3 tells us that x = 0.

All we have left to determine is the vaue of z.

Borrowing Equation 3, from above, we get x - 2 y - 3 z = 0, and into this equation we substitute x = 0 and y = 3: 0 -2(3) - 3z = 0.

Thus, -3z = 6, and z = -2.

The solution set is (0, 3, -2). You should check this by substitution.

3 0

4 years ago

Pls help me on this question

Maksim231197 [3]

The answer is selections 1 and 4

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coldgirl [10]

Answer:

Step-by-step explanation:

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