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3 years ago

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On the coordinate grid, the graph of y = RootIndex 3 StartRoot negative x minus 1 EndRoot is shown. It is a reflection and trans

lation of y = RootIndex 3 StartRoot x EndRoot. On a coordinate plane, a cube root function goes through (negative 2, 1), has an inflection point at (negative 1, 0), and goes through (7, negative 2). What is the range of the graphed function? {x |-2 < x < 2} {y |-2 < y < 2} {x | x is a real number} {y | y is a real number}

1 answer:

Thepotemich [5.8K]3 years ago

5 0

Answer:

{y | y is a real number}

Step-by-step explanation:

Just did the test on edgenuity

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A student paid 256.37 in school fees how much has he paid to the nearest hundred <br><br>

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300

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The square of a number is increased by 27 and the result is 148. Find all possible solutions for the number.

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A rectangle is placed around a semicircle as shown below. The length of the rectangle is 8 cm. Find the area of the shaded regio

icang [17]

Answer:

Area of the shaded region = 6.88 cm²

Step-by-step explanation:

Area of the shaded region = Area of the rectangle - Area of the semicircle

Area of the rectangle = Length × Width

Length of the rectangle = AD = 8 cm

Width of the rectangle = OP = 4 cm [OP, OB and OC are the radii of the circle which are equal in measures]

Area of the rectangle = 8 × 4 = 32 cm²

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3 years ago

Evaluate 32 + 12 = ( 6 – 3) x 8.

Brrunno [24]

Hello there!

(Equation #1) 32 + 12 = 44

(Equation #2) 6 - 3 = 3, 3 x 8 = 24

These both equations don't have the same answer, even though it says that they are equal with each other, if you could correct these these 2 equations I am willing to answer again.

Hope this helped!!

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3 years ago

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What two rational expressions sum to 2x+3/x^2-5x+4

Anni [7]

Answer:

$\frac{2x + 3}{(x- 1)(x - 4)} = \frac{-5}{3(x- 1)} + \frac{11}{3(x - 4)}$

Step-by-step explanation:

Given the rational expression: $\frac{2x + 3}{x^2 - 5x + 4}$ , to express this in simplified form, we would need to apply the concept of partial fraction.

Step 1: factorise the denominator

$x^2 - 5x + 4$

$x^2 - 4x - x + 4$

$(x^2 - 4x) - (x + 4)$

$x(x - 4) - 1(x - 4)$

$(x- 1)(x - 4)$

Thus, we now have: $\frac{2x + 3}{(x- 1)(x - 4)}$

Step 2: Apply the concept of Partial Fraction

Let,

$\frac{2x + 3}{(x- 1)(x - 4)}$ = $\frac{A}{x- 1} + \frac{B}{x - 4}$

Multiply both sides by (x - 1)(x - 4)

$\frac{2x + 3}{(x- 1)(x - 4)} * (x - 1)(x - 4)$ = $(\frac{A}{x- 1} + \frac{B}{x - 4}) * (x - 1)(x - 4)$

$2x + 3 = A(x - 4) + B(x - 1)$

Step 3:

Substituting x = 4 in $2x + 3 = A(x - 4) + B(x - 1)$

$2(4) + 3 = A(4 - 4) + B(4 - 1)$

$8 + 3 = A(0) + B(3)$

$11 = 3B$

$\frac{11}{3} = B$

$B = \frac{11}{3}$

Substituting x = 1 in $2x + 3 = A(x - 4) + B(x - 1)$

$2(1) + 3 = A(1 - 4) + B(1 - 1)$

$2 + 3 = A(-3) + B(0)$

$5 = -3A$

$\frac{5}{-3} = \frac{-3A}{-3}$

$A = -\frac{5}{3}$

Step 4: Plug in the values of A and B into the original equation in step 2

$\frac{2x + 3}{(x- 1)(x - 4)} = \frac{A}{x- 1} + \frac{B}{x - 4}$

$\frac{2x + 3}{(x- 1)(x - 4)} = \frac{-5}{3(x- 1)} + \frac{11}{3(x - 4)}$

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3 years ago