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ale4655 [162]
3 years ago
11

2.

Mathematics
1 answer:
Vadim26 [7]3 years ago
3 0

Answer:

The answer is A. f(x) ≥ 2x2 + 4x or C. f(x) ≤ 2x2 + 4x

Step-by-step explanation:

Vertex:  

( − 1 , − 2 )

Focus:  

( − 1 , − 15 /8 )

Axis of Symmetry:  

x

=

−

1

Directrix:  

y

=

−

17

8

x

y

−

3

6

−

2

0

−

1

−

2

0

0

1

6

Vertex:  

( − 1 , − 2 )

Focus:  

( − 1 , − 15 /8 )

Axis of Symmetry:  

x

=

−

1

Directrix:  

y

=

−

17

8

x

y

−

3

6

−

2

0

−

1

−

2

0

0

1

6

Vertex:  

( 1 , 2 )

Focus:  

( 1 , 15 /8 )

Axis of Symmetry:  

x

=

1

Directrix:  

y

=

17

8

x

y

−

1

−

6

0

0

1

2

2

0

3

−

6

Vertex:  

( − 1 , − 2 )

Focus:  

( − 1 , − 15 /8 )

Axis of Symmetry:  

x

=

−

1

Directrix:  

y

=

−

17

8

x

y

−

3

6

−

2

0

−

1

−

2

0

0

1

6

Vertex:  

( 1 , 2 )

Focus:  

( 1 , 15 /8 )

Axis of Symmetry:  

x

=

1

Directrix:  

y

=

17

8

x

y

−

1

−

6

0

0

1

2

2

0

3

−

6

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Answer: f(-3) = 278, f(-4) = 232

Step-by-step explanation:

f(x) = 4x³ - 6x² - 144x + 8

f(-3) = 4(-3)³ - 6(-3)² - 144(-3) + 8

f(-3) = 4(-27) - 6(9) - 144(-3) + 8

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f(-4) = 4(-64) - 6(16) - 144(-4) + 8

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Answer:

Business-equipment loan, A=$15000

Small-business loan, B=$24000

home-equity loan, C= $39000

Step-by-step explanation:

Let the Business-equipment loan at an interest rate of​ 11%=A

Let the​ small-business loan was at an interest rate of 5​% =B

Let her​ home-equity loan was at an interest rate of 4.5​%.=C

Since her total Loan=$78000

A+B+C=78000....(I)

Simple Interest = (P X R X T)/100

Since the Time, T=1 year

Interest on A = 0.11A

Interest on B = 0.05B

Interest on C = 0.045C

The total simple interest due on the loans in one year was ​$4605.

0.11A+0.05B+0.045C=4605....(II)

The annual simple interest on the​ home-equity loan was ​$105 more than the interest on the​ business-equipment loan.

0.045C = 0.11A +105...(III)

We proceed to solve the simultaneous equations.

A+B+C=78000....(I)

0.11A+0.05B+0.045C=4605....(II)

0.045C = 0.11A +105...(III)

From (III), 0.045C = 0.11A +105

Substitute 0.045C = 0.11A +105 into (II).

0.11A+0.05B+0.11A +105=4605

0.22A+0.05B=4500

From (III),

C=\frac{22A}{9}+\frac{7000}{3}

Substitute into (I)

A+B+\frac{22A}{9}+\frac{7000}{3}=78000

\frac{31A}{9}+B=78000-\frac{7000}{3}

\frac{31A}{9}+B=\frac{227000}{3}

\frac{31A+9B}{9}=\frac{227000}{3}

3(31A+9B)=9 X 227000

31A+9B=681000

0.22A+0.05B=4500 (Multiply by 9)

31A+9B=681000 (Multiply by 0.05)

1.98A+0.45B=40500

1.55A+0.45B=34050

Subtracting

0.43A=6450

A=$15000

From (III)

C=\frac{22A}{9}+\frac{7000}{3}

C=\frac{22X15000}{9}+\frac{7000}{3} =39000

C=$39000

from (I)

A+B+C=78000

15000+B+39000=78000

B=$24000

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