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2 years ago

What is the vertex of the absolute value function defined by f(x) = (x + 2) + 4?

Mathematics

1 answer:

SSSSS [86.1K]2 years ago

8 0

Answer:

Step-by-step explanation:

maximum point should be 4

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Expand. Your answer should be a polynomial in standard form. (x2+4)(x2−6)=(x^2+4)(x^2-6)=(x2+4)(x2−6)=

Dafna11 [192]

Answer:

x^4 -2x^2 -24

Step-by-step explanation:

(x^2+4)(x^2−6)

FOIL

First: x^2*x^2 = x^4

Outer: x^2 *-6 = -6x^2

Inner: x^2 *4 = 4x^2

Last: 4*-6 = -24

Add them together

x^4 -6x^2 +4x^2 -24

Combine like terms

x^4 -2x^2 -24

This is in standard form since the powers decrease

8 0

2 years ago

Read 2 more answers

Antonio is going to the carnival. He can spend no more than $20.00. if the ticket to in is $7.50, and each ride cost $1.25, how

fgiga [73]

$20.00 ≥ 7.50+1.25x
subtract 7.50 from both sides
20-7.50=12.50
12.50=1.25x
12.50/1.25=10
x= 10 rides

6 0

3 years ago

Please help! Find the area and the perimeter of the shaded regions below. Give your answer as a completely simplified exact valu

Nadya [2.5K]

Answer:

$The \ area \ of \ the \ figure = 4\dfrac{1}{2} \cdot \pi + 18 \ cm^2$

Step-by-step explanation:

Given that the figure is made up of portion of a square and a semicircle, we have;

BC ≅ AB = 6 cm

The area of semicircle BC with radius BC/2 = 3 is 1/2×π×r² = 1/2×π×3² = 4.5·π cm²

Triangle ABC = 1/2 × Area of square from which ABC is cut

The area of triangle ABC = 1/2×Base ×Height = 1/2×AB×BC = 1/2×6×6 = 18 cm²

The area of the figure = The area of semicircle BC + The area of triangle ABC

The area of the figure = 4.5·π cm² + 18 cm² = $\dfrac{9 \cdot \pi +36}{2} \ cm^2$ .

5 0

3 years ago

Write five names for -73

Deffense [45]

Negative 73
drop of 73
minus 73
below 73
taking away 73

6 0

2 years ago

A/ab-bsquare + b/ab-asquare?

lord [1]

Answer:

$\dfrac{(a+b)}{ab}$

Step-by-step explanation:

The given expression is :

$\dfrac{a}{ab-b^2}+\dfrac{b}{ab-a^2}$

It can be solved as follows :

$\dfrac{a}{ab-b^2}+\dfrac{b}{ab-a^2}=\dfrac{a}{b(a-b)}+\dfrac{b}{a(b-a)}\\\\=\dfrac{a}{b(a-b)}+\dfrac{b}{-a(-b+a)}\\\\=\dfrac{1}{a-b}(\dfrac{a}{b}-\dfrac{b}{a})\\\\=\dfrac{a^2-b^2}{ab(a-b)}\\\\=\dfrac{(a-b)(a+b)}{ab(a-b)}\\\\=\dfrac{(a+b)}{ab}$

So, the solution of the given expression is equal to $\dfrac{(a+b)}{ab}$ .

7 0

2 years ago