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2 years ago

Which part of figure A is misleading ?

Mathematics

1 answer:

Marta_Voda [28]2 years ago

7 0

Answer:

Step-by-step explanation:

It looks like figure A shows lower scores since it doesn't start at 0

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Write 0.87 as a fraction

Deffense [45]

0.87 = 87/100 <=== this is the fraction

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3 years ago

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elaina wrote two correct variable expressions to represent the phrase “one-fourth of a number.” which variable expressions could

antiseptic1488 [7]

One-fourth of a number

1/4n and n/4

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3 years ago

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Find the zeros of the function.<br> f(x) = 3x2 – 8x – 16

Vlad1618 [11]

Answer:

amusing you mean 3x^2-8x-16 the answer is when x is equal to -4/3 and 4

Step-by-step explanation:

3x^2-8x-16 ------> (3x+4)(x-4)

3x+4=0 ----> 3x=-4 ----> x=-4/3

x-4=0 ----> x=4

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3 years ago

The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t

shtirl [24]

Answer:

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

Step-by-step explanation:

As the given Augmented matrix is

$\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 1 :

$r_{1}$ ↔ $r_{1} - r_{2}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 2 :

$r_{3}$ ↔ $r_{3} - 8r_{1}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]$

Step 3 :

$r_{2}$ ↔ $\frac{r_{2}}{7}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]$

Step 4 :

$r_{1}$ ↔ $r_{1} + 14r_{2}$ , $r_{3}$ ↔ $r_{3} - 124r_{2}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]$

Step 5 :

$r_{3}$ ↔ $\frac{r_{3}. 7}{254}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]$

Step 6 :

$r_{1}$ ↔ $r_{1} - 4r_{3}$ , $r_{2}$ ↔ $r_{2} + \frac{1}{7} r_{3}$

$\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]$

∴ we get

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

6 0

2 years ago

If one factor of x2 + 2x - 24 is (x+6), what is the other factor?

steposvetlana [31]

Answer:

( x - 4)

Step-by-step explanation:

x² + 2x - 24

From here, you must find the two factors of -24 (the first and third terms' coefficients multiplied together) that add up to 2 (the middle terms coefficient)

( x - 4) ( x + 6)

4 0

2 years ago

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