If h(x) = x – 7 and g(x) = x2, which expression is equivalent to mc022-1.jpg?

2 answers:

8 0

the answer:
the full question is as follow:

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

A (5 – 7)2,
B (5)2 – 7,
C (5)2(5 – 7),
D (5 – 7)x2

first, the main rule of the product between two functions g and h is

g(x)*h(x)= (g*h)(x)

h(x) = x – 7 and g(x)=x², so (g*h)(x)= g(x)*h(x)=[x²][x – 7 ]= x^3 -7x²

(g*h)(x)= x^3 -7x², therefore, (g*h)(5)= 5^3 -7*5² = -50 = (5)2(5 – 7)

finally, the answer is C (5)2(5 – 7),

Yuri [45]4 years ago

5 0

Answer:

A. (5-7)2

Step-by-step explanation:

I took an educational guess on edgen. and got this right.

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cluponka [151]

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3 0

3 years ago

Solve the equation; 3a/4-2a/3=7/4

DiKsa [7]

Answer: " a = 21 " .

_____________________

Step-by-step explanation:

Given: 3a/4-2a/3 = 7/4 ; Solve for "a" ;

Rewrite as: (3a/4) - (2a/3) = (7/4) ;

Now, for each of the three (3) "denominator values" in "fraction form" within the equation given:

→ Find the "LCD" ["Least Common Denominator"]:
The denominators are: 4, 3, and 4 ;

that is: "3" and "4" ;
____

To find the LCD: First; multiply the denominators: "4 * 3 = 12" .
So; the value "12" could be the LCD; so, the value for the LCD is no greater than "12" ; however, there could be a smaller value.
To determine the LCD:
List the multiples of the given denominators:
____
3: 3, 6, 9, 12, 15 .... ;
4: 4, 8, 12, 16... ;
____
We find that "12" is, in fact, the LCD of "3" and 4:
____
We can multiply each side of the equation by "12" ; to eliminate the "fractional values" :
____
$12*[\frac{3a}{4} - \frac{2a}{3}] = 12*[{\frac{7}{4}]$
____
Note the "distributive property" of multiplication:
→ a(b+c) = ab + ac ;

____

As such:
$12*[\frac{3a}{4} - \frac{2a}{3}] = 12*[{\frac{7}{4}]$ ;

____
Let us start with the "left-hand side" of the equation:
____
$12*[\frac{3a}{4} - \frac{2a}{3}]$ ;

= $[12*\frac{3a}{4}] + [-12 * \frac{2a}{3}]$ ;
= $[12*\frac{3a}{4}] - [12 * \frac{2a}{3}] ;$
____
Note: " $[12*\frac{3a}{4}]$ " ;

$= \frac{12}{1} * \frac{3a}{4}$ ;

→ The "12" cancels to a "3" ; and the "4" cancels to a "1" ;
since: "12÷4 = 3" ; and since: "4÷4 = 1" ;
→ and we can rewrite the "left-hand-side" expression as:
→ " $\frac{3}{1} * \frac{3a}{1}$ " ; which we can simplify as:

→ "3 * 3a" ; which we can simplify as: " 9a " .
then we have: " [12 * $\frac{2a}{3}$ ] " ;

which equals:
= " $\frac{12}{1} * \frac{2a}{3}$ " ;
Note: The "12" cancels out to a "4"; & the "3" cancels out to a "1" ;

→ {since: "(12 ÷ 3 = 4)"; & since: "(3 ÷ 3 = 1)" ;
____
→ And we can rewrite the expression as:
→ " $\frac{4}{1} *\frac{2a}{1}$ " ; which we can simplify as:
→ " 4 * 2a " ; which we can simplify/calculation as: " 8a " ;
Now, we can rewrite the expression of the "left-hand side"
of the equation as:
____
→ " [9a] − [8a] " ; (don't forget to carry down the "minus sign"!) ;
which we can simplify/calculate to get:
→ " [9a − 8a] " ; which we can further simplify/calculate;

→ to get:
→ " 1a " ; or: "a" —the value for which we wish to solve!
____
Now, let us examine the "right-hand side" of the equation:
____

→ " $\frac{12}{1} * \frac{7}{4}$ " ;
Note: The "12" cancels out to a "3" ; & the "4" cancels out to a "1" ;
→ {Since: "12 ÷4 = 3 " ; & since: "4 ÷ 4 = 1 "} ;

And we can rewrite the expression as:
→ " $\frac{3}{1} * \frac{7}{1}$ " ; which we can simplify as:
→ " 3 * 7 " ; which can simplify/calculate to get: " 21" ;
⇒ Now, let us rewrite the equation; by using our simplified values for both the "left-hand side" and the "right-hand side" of the equation; to solve for "a" :
⇒ a = 21 ;