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2 years ago

If h(x) = x – 7 and g(x) = x2, which expression is equivalent to (g•h)(5)?

Mathematics

2 answers:

maks197457 [2]2 years ago

4 0

Answer:

Step-by-step explanation:

timofeeve [1]2 years ago

3 0

Answer:

(g · h)(5) = -50

Step-by-step explanation:

Step 1: Define

h(x) = x - 7

g(x) = x²

Step 2: Find (g · h)(x)

(g · h)(x) = x²(x - 7)

(g · h)(x) = x³ - 7x²

Step 3: Find (g · h)(5)

(g · h)(5) = 5³ - 7(5)²

(g · h)(5) = 125 - 7(25)

(g · h)(5) = 125 - 175

(g · h)(5) = -50

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2 years ago

You are skateboarding at a pace of 30 meters every 5 seconds. your friend is in-line skating at a pace of 9 meters every 2 secon

AnnyKZ [126]

In order to graph the relationship, we will need to write the expression as the equation of a straight line as shown:

d = mt + b

d is the distance covered

t is the time taken

m is the speed

If you are skateboarding at a pace of 30 meters every 5 seconds. your friend is in-line skating at a pace of 9 meters every 2 seconds, this can be written as (5, 30) and (2, 9)

Get the slope of the line:

m = (9-30)/(2-5)

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m = 7

Substitute m = 7 and the coordinate (2, 9) into the equation y = mt + b

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The required equation to plot will be expressed as y = 7t - 5

Plot the required graph

Learn more here: brainly.com/question/17003809

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2 years ago

The graph of y = x 2 has been translated 7 units to the left. The equation of the resulting parabola is _____.

strojnjashka [21]

To move the function to the left you have to increase x inside the function
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6 0

2 years ago

Read 2 more answers

Please help me on this problem

Anna71 [15]

Answer:

The pairs of integer having two real solution for $ax^{2} -6x+c = 0$ are

$a = -4, c = 5$
$a = 1, c = 6$
$a = 2, c = 3$
$a = 3, c = 3$

Step-by-step explanation:

Given

$ax^{2} -6x+c = 0$

Now we will solve the equation by putting all the 6 pairs so we get the following

$-3x^{2} -6x-5 = 0$ for $a = -3 , c=-5$

$-4x^{2} -6x+5 = 0$ for $a = -4 , c=5$

$1x^{2} -6x+6 = 0$ for $a = 1 , c=6$

$2x^{2} -6x+3 = 0$ for $a = 2 , c=3$

$3x^{2} -6x+3 = 0$ for $a = 3 , c=3$

$5x^{2} -6x+4 = 0$ for $a = 5 , c=4$

The above all are Quadratic equations inn general form $ax^{2} +bx+c=0$

where we have a,b and c constant values

So for a real Solution we must have

$Disciminant , b^{2} -4\timesa\timesc \geq 0$

for $a = -3 , c=-5$ we have

$Discriminant =-24$ which is less than 0 ∴ not a real solution.

for $a = -4 , c=5$ we have

$Discriminant = 116$ which is greater than 0 ∴ a real solution.

for $a = 1 , c=6$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 2 , c=3$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 3 , c=3$ we have

$Discriminant =0$ which is equal to 0 ∴ a real solution.

for $a = 5 , c=4$ we have

$Discriminant =-44$ which is less than 0 ∴ not a real solution.

7 0

2 years ago