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2 years ago

Use synthetic substitution to find f (2) and f (-1) for the function. f (x) = x3 – x2 – 2x + 3

Mathematics

1 answer:

kotegsom [21]2 years ago

3 0

Answer:

f(2)=7

f(-1)=+1

Step-by-step explanation:

f (2)= 2³-2²+3 =8-4+3=7

f (-1)= (-1)³-(-1)²+3= -1- (+1) +3= -1-1+3=+1

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Given f(x) = 2x + 3 and g(x) = x - 9, what is f(x) + g(x)?

Orlov [11]

Solution:

f(x) = 2x + 3
g(x) = x - 9

f(x) + g(x)
= (2x + 3) + (x - 9)
= 2x + x + 3 - 9
= 3x - 6

6 0

2 years ago

A triangle with sides 11m , 13m and 18m is a right triangle. A True B False

Julli [10]

$\bold{\huge{\pink{\underline{ Solution}}}}$

$\bold{\underline{ Given :-}}$

A triangle with sides 11m, 13m and 18m

$\bold{\underline{ To \: Find :-}}$

We have to check it whether it is right angled triangle or not?

$\bold{\underline{ Let's \: Begin :-}}$

$\sf{\red{ In \:right \:angled\ : triangle, }}$

According to the Pythagoras theorem, The sum of the squares of perpendicular height and the square of the base of the triangle is equal to the square of hypotenuse that is sum of the squares of two small sides equal to the square of longest side of the triangle.

We imply it in the given triangle ,

$\sf{\red{ ( Perpendicular)² + (Base)² = (Hypotenuse)²}}$

$\sf{(AB)² + (BC)² = (AC)²}$

$\sf{ (11)² + (13)² = (18)² }$

$\sf{ 121 + 169 ≠ 324 }$

$\sf{ 290 ≠ 324 }$

From Above we can conclude that,

The sum of the squares of two small sides that is perpendicular height and base is not equal to the square of longest side that is Hypotenuse

$\sf{\blue{ Hence,\: Your\: answer \:is \:false }}$

5 0

1 year ago

25 divided by something = 3 remainder 4

NeX [460]

7 would be the number cause;

25 ÷ 7 = 3 r4

hope that helps :D

8 0

3 years ago

Two friends sold many pieces of furniture and made $1550 during their garage sale. They had fifteen more $10 bills than $50 b

Novosadov [1.4K]

Answer:

The number of $ 10 bills is 37, the number of $ 20 is 6 and the number of $ 50 bills is 22.

Step-by-step explanation:

Let n₁ = number of $ 10 bills,

Let n₂ = number of $ 20 bills and

Let n₃ = number of $ 50 bills

Given that 10n₁ + 20n₂ + 50n₃ = 1550 (1)

Also, since we have fifteen more $10 bills than $50 bills, n₁ = n₃ + 15 (2)

and we have 4 more than three times as many $20 bills as $50 bills. n₃ = 3n₂ + 4. (3)

substituting equations (2) and (3) into (1), we have

10(n₃ + 15) + 20n₂ + 50n₃ = 1550

expanding the bracket, we have

10n₃ + 150 + 20n₂ + 50n₃ = 1550

collecting like terms, we have

60n₃ + 150 + 20n₂ = 1550

inserting equation (3), we have

60(3n₂ + 4) + 150 + 20n₂ = 1550

expanding the bracket, we have

180n₂ + 240 + 150 + 20n₂ = 1550

collecting like terms, we have

200n₂ + 390 = 1550

subtracting 390 from both sides, we have

200n₂ = 1550 - 390

200n₂ = 1160

dividing both sides by 200, we have

n₂ = 1160/200

n₂ = 5.8

n₂ ≅ 6 since it cannot be a fraction.

Substituting this into (3), we have

n₃ = 3n₂ + 4 = 3(6) + 4 = 18 + 4 = 22

substituting n₃ into (2), we have

n₁ = n₃ + 15 = 22 + 15 = 37

So, the number of $ 10 bills is 37, the number of $ 20 is 6 and the number of $ 50 bills is 22.

3 0

2 years ago

What is the solution for the following equation 3(2x+3)=2(3x+4)?

ki77a [65]

Answer:

<h2>no solution, the equation is false</h2>

Step-by-step explanation:

what is the solution for the following equation 3(2x+3)=2(3x+4)?

3 * (2x + 3) = 2 * (3x + 4)

6x + 9 = 6x + 8

6x - 6x = -9 + 8

0 = ±1

the equation is false

7 0

1 year ago

Read 2 more answers