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lana [24]
3 years ago
9

Terrell wrote that 5 – 4.2 = 8. Which statement about his answer is true? CLEAR CHECK 8 is wrong because a number gets smaller,

not larger, when you take 4.2 from it. 8 is right because 50 – 42 = 8. 8 is right because 4.2 + 8 = 5. 8 is wrong because you cannot subtract a decimal from a whole number.
Mathematics
1 answer:
SVEN [57.7K]3 years ago
5 0

Answer:

a) is wrong because a number gets smaller, not larger, when you take 4.2 from it

Step-by-step explanation:

Given

5 - 4.2 = 8

Required

Determine which of the options is true

First, we need to solve for 5 - 4.2

5 - 4.2 = 0.8

Compare this answer with Terell's answer (8)

We observe that: 0.8 \neq 8

This implies that; Terell is wrong

The two options that support the answer being wrong are (a) & (d)

Analyzing (a):

5 should get smaller when 4.2 is subtracted from it.

This isn't the case in Terell's solution.

Hence, (a) is true

Analyzing (d):

d is wrong because a decimal can be subtracted from a whole number

<em>Hence, (a) answers the question</em>

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Identify the interval on which the quadratic function is positive.
Alenkasestr [34]

Answer:

\textsf{1. \quad Solution:  $1 < x < 4$,\quad  Interval notation:  $(1, 4)$}

\textsf{2. \quad Solution:  $-2 < x < 4$,\quad  Interval notation:  $(-2, 4)$}

Step-by-step explanation:

<h3><u>Question 1</u></h3>

The intervals on which a <u>quadratic function</u> is positive are those intervals where the function is above the x-axis, i.e. where y > 0.

The zeros of the <u>quadratic function</u> are the points at which the parabola crosses the x-axis.  

As the given <u>quadratic function</u> has a negative leading coefficient, the parabola opens downwards.   Therefore, the interval on which y > 0 is between the zeros.

To find the zeros of the given <u>quadratic function</u>, substitute y = 0 and factor:

\begin{aligned}y&= 0\\\implies -7x^2+35x-28& = 0\\-7(x^2-5x+4)& = 0\\x^2-5x+4& = 0\\x^2-x-4x+4& = 0\\x(x-1)-4(x-1)&= 0\\(x-1)(x-4)& = 0\end{aligned}

Apply the <u>zero-product property</u>:

\implies x-1=0 \implies x=1

\implies x-4=0 \implies x=4

Therefore, the interval on which the function is positive is:

  • Solution:  1 < x < 4
  • Interval notation:  (1, 4)

<h3><u>Question 2</u></h3>

The intervals on which a <u>quadratic function</u> is negative are those intervals where the function is below the x-axis, i.e. where y < 0.

The zeros of the <u>quadratic function</u> are the points at which the parabola crosses the x-axis.  

As the given <u>quadratic function</u> has a positive leading coefficient, the parabola opens upwards.   Therefore, the interval on which y < 0 is between the zeros.

To find the zeros of the given <u>quadratic function</u>, substitute y = 0 and factor:

\begin{aligned}y&= 0\\\implies 2x^2-4x-16& = 0\\2(x^2-2x-8)& = 0\\x^2-2x-8& = 0\\x^2-4x+2-8& = 0\\x(x-4)+2(x-4)&= 0\\(x+2)(x-4)& = 0\end{aligned}

Apply the <u>zero-product property</u>:

\implies x+2=0 \implies x=-2

\implies x-4=0 \implies x=4

Therefore, the interval on which the function is negative is:

  • Solution:  -2 < x < 4
  • Interval notation:  (-2, 4)
3 0
1 year ago
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