All categories

3 years ago

Given f(x)=22x+12, find f(7)

Mathematics

2 answers:

krek1111 [17]3 years ago

8 0

166

Plug in 7 for x

I need more letters

Zarrin [17]3 years ago

4 0

Substitute x for 7.

f(x)=22x+12

f(7)=22(7)+12

Multiply first, then complete addition.

f(7)=154+12

f(7)=166

You might be interested in

--(4+h) = 3h<br> What is H

butalik [34]

H= -1

Distribute:

-(4+h)= -4 - h

-4 - h= 3h

-4 +h = 3h + h

-4= 4h

-1 = h

7 0

3 years ago

What is the equation of the line

Sphinxa [80]

x = 0

................

7 0

3 years ago

Read 2 more answers

Which inequality best represents the situation in which you must be at least 42 inches tall to ride the roller coaster

Romashka-Z-Leto [24]

$42 \leq x$

6 0

3 years ago

in these triangles, side AB is congruent to side DF, and side BC is congruent to side FG. Determine the values of x and y

vlada-n [284]

I added a screenshot with the complete question

Answer:
x = 3
y = 9

Explanation:
1- getting the value of x:
We are given that:
side AB is congruent to side DF. This means that:
AB = DF
3(2x+10) = 12x + 12
6x + 30 = 12x + 12
12x - 6x = 30 - 12
6x = 18
x = 18/6
x = 3

2- getting the value of y:
We are given that:
side BC is congruent to side FG. This means that:
BC = FG
2y + 12 = 2(2y-3)
2y + 12 = 4y - 6
4y - 2y = 12 + 6
2y = 18
y = 18/2
y = 9

Hope this helps :)

5 0

3 years ago

The polynomial of degree 5, P ( x ) has leading coefficient a=1, has roots of multiplicity 2 at x = 3 and x = 0 , and a root of

ser-zykov [4K]

Answer:

$p_{5} (t) = x^{5} - 5\cdot x^{4} + 3\cdot x^{3} +9\cdot x^{2}$ , for $r_{1} = 0$

Step-by-step explanation:

The general form of quintic-order polynomial is:

$p_{5}(t) = a\cdot x^{5} + b\cdot x^{4} + c\cdot x^{3} + d\cdot x^{2} + e \cdot x + f$

According to the statement of the problem, the polynomial has the following roots:

$p_{5} (t) = (x - r_{1})\cdot (x-3)^{2}\cdot x^{2} \cdot (x+1)$

Then, some algebraic handling is done to expand the polynomial:

$p_{5} (t) = (x - r_{1}) \cdot (x^{3}-6\cdot x^{2}+9\cdot x) \cdot (x+1)\\p_{5} (t) = (x - r_{1}) \cdot (x^{4}-5\cdot x^{3} + 3 \cdot x^{2} + 9 \cdot x)$

$p_{5} (t) = x^{5} - (5+r_{1})\cdot x^{4} + (3 + 5\cdot r_{1})\cdot x^{3} +(9-3\cdot r_{1})\cdot x^{2} - 9 \cdot r_{1}\cdot x$

If $r_{1} = 0$ , then:

$p_{5} (t) = x^{5} - 5\cdot x^{4} + 3\cdot x^{3} +9\cdot x^{2}$

5 0

2 years ago