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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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17. A gardener is installing a fence around his garden. Let x represent the width of

Serhud [2]

Hey there!

Answer:

B. 3x + 4 ft.

Step-by-step explanation:

Remember that:

Perimeter of a rectangle: P = 2l + 2w

Given:

P = 8x + 8

w = x.

Plug the given information into the equation:

8x + 8 = 2l + 2x

Subtract 2x from both sides:

6x + 8 = 2l

Divide both sides by 2:

(6x + 8)/2 = 2l/2

3x + 4 = l.

Therefore, the length of the garden is:

l = 3x + 4 ft.

6 0

4 years ago

Read 2 more answers

For f(x)=4sin(x2) between x=0 and x=3, find the coordinates of all intercepts, critical points, and inflection points to two dec

telo118 [61]

Answer:

Intercepts:

x = 0, y = 0

x = 1.77, y = 0

x = 2.51, y = 0

Critical points:

x = 1.25, y = 4

x = 2.17 , y = -4

x = 2.8, y = 4

Inflection points:

x = 0.81, y = 2.44

x = 1.81, y = -0.54

x = 2.52, y = 0.27

Step-by-step explanation:

We can find the intercept by setting f(x) = 0

$4sin(x^2) = 0$

$sin(x^2) = 0$

$x^2 = n\pi$ where n = 0, 1, 2,3, 4, 5,...

$x = \sqrt(n\pi)$

Since we are restricting x between 0 and 3 we can stop at n = 2

So the function f(x) intercepts at y = 0 and x:

x = 0

x = 1.77

x = 2.51

The critical points occur at the first derivative = 0

$f^{'}(x) = 4cos(x^2)2x = 8xcos(x^2) = 0$

$xcos(x^2) = 0$

$x = 0$ or

$cos(x^2) = 0$

$x^2 = \frac{\pi}{2} + n\pi$ where n = 0, 1, 2, 3

$x = \sqrt{\pi(n+1/2)}$

Since we are restricting x between 0 and 3 we can stop at n = 2

So our critical points are at

x = 1.25, $y = f(1.25) = 4sin(1.25^2) = 4$

x = 2.17 , $y = f(2.17) = 4sin(2.17^2) = -4$

x = 2.8, $y = f(2.8) = 4sin(2.8^2) = 4$

For the inflection point, we can take the 2nd derivative and set it to 0

$f^[''}(x) = 8(cos(x^2) - xsin(x^2)2x) = 8cos(x^2) - 16x^2sin(x^2) = 0$

$cos(x^2) = 2x^2sin(x^2)$

$tan(x^2) = \frac{1}{2x^2}$

We can solve this numerically to get the inflection points are at

x = 0.81, $y = f(0.81) = 4sin(0.81^2) = 2.44$

x = 1.81, $y = f(1.81) = 4sin(1.81^2) = -0.54$

x = 2.52, $y = f(2.52) = 4sin(2.52^2) = 0.27$

3 0

3 years ago

In his last game the Rams” quarterback threw 18

Marrrta [24]

the answer is 15 complete passes

8 0

2 years ago

Read 2 more answers

Abc bookstore sells new books, n, for $12, used books, u, for $8, and magazines, m, for $5 each. The store earned $340 revenue l

poizon [28]

Answer:

The number of new books sold = 10

The number of used books sold (u) = 15

The number of magazines sold (m) = 20

Step-by-step explanation:

Let us assume the number of new books = n

So, the number of used books sold (u) = New books sold + 5 = n + 5

Also, the number of magazines sold (m) = 2 x ( Number of new books sold)

= 2 n

⇒ u = n + 5, m = 2 n

Here, the cost of each new book n- = $12

So, the cost of n new books = n x ($12) = 12 n

the cost of each used book u = $8

So, the cost of u = (n+ 5) used books = n+5 x ($8) = 8 n + 40

the cost of each magazine m = $5

So, the cost of m = (2n ) magazines = 2n x ($5) = 10 n

Also, total earnings = $340

⇒ 12 n + 8n +40 + 10 n = 340

or, 30 n = 300

or,n = 300/30 = 10

Hence the number of new books sold = n = 10

The number of used books sold (u) = n + 5 = 15

The number of magazines sold = m = 2 n = 20

4 0

3 years ago

At any point in time, there could be bicycles, tricycles, and

Aleksandr-060686 [28]

Answer:

There may be 1 or 3 tricycles in the parking lot.

Step-by-step explanation:

Since at any point in time, there could be bicycles, tricycles, and cars in the school parking lot, and today, there are 53 wheels in total, if there are 15 bicycles, tricycles, and cars in total, to determine how many tricycles could be in the parking lot, the following calculation must be performed:

13 x 4 + 1 x 3 + 1 x 2 = 57

11 x 4 + 1 x 3 + 3 x 2 = 53

10 x 4 + 3 x 3 + 2 x 2 = 53

8 x 4 + 5 x 3 + 2 x 2 = 51

10 x 2 + 1 x 3 + 4 x 4 = 39

9 x 3 + 1 x 2 + 5 x 4 = 49

Therefore, there may be 1 or 3 tricycles in the parking lot.

6 0

3 years ago