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PolarNik [594]
3 years ago
14

A student took 60 minutes to answer a combination of 20 multiple choice and extended response choice questions she took two minu

tes to anwer each multiple choice and 6 minutes to anwaer each extended response.
A) write a system of equations to model the relationship between the number of multiple choice questions M and number of extended response questions R
B) how many of each type of question was on the test
Mathematics
2 answers:
Ray Of Light [21]3 years ago
8 0
20 = m + r
60 = 2m + 6r

Using substitution, you would subtract R from both sides in the first equation.

20 - r = m

Substitute (20 - r) for m in the second equation.

60 = 2(20 - r) + 6r

Distribute the 2

60 = 40 - 2r +6r

Combine like terms.

20 = 4r

Divide both sides by 4

5 = r

Now put R back into the first equation.

20 - 5 = m

15 = m

There were 15 multiple choice and 5 extended response questions.
frez [133]3 years ago
4 0
The average rate of change of g(x) between x = 4 and x = 7 is . Which statement must be true?

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Choose the equation that represents the solutions of 0 = 0.25x2 - 8x.
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3 0
3 years ago
Read 2 more answers
1) Determine the discriminant of the 2nd degree equation below:
Aleksandr-060686 [28]

\LARGE{ \boxed{ \mathbb{ \color{purple}{SOLUTION:}}}}

We have, Discriminant formula for finding roots:

\large{ \boxed{ \rm{x =  \frac{  - b \pm \:  \sqrt{ {b}^{2}  - 4ac} }{2a} }}}

Here,

  • x is the root of the equation.
  • a is the coefficient of x^2
  • b is the coefficient of x
  • c is the constant term

1) Given,

3x^2 - 2x - 1

Finding the discriminant,

➝ D = b^2 - 4ac

➝ D = (-2)^2 - 4 × 3 × (-1)

➝ D = 4 - (-12)

➝ D = 4 + 12

➝ D = 16

2) Solving by using Bhaskar formula,

❒ p(x) = x^2 + 5x + 6 = 0

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - 5\pm  \sqrt{( - 5) {}^{2} - 4 \times 1 \times 6 }} {2 \times 1}}}

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - 5  \pm  \sqrt{25 - 24} }{2 \times 1} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - 5 \pm 1}{2} }}

So here,

\large{\boxed{ \rm{ \longrightarrow \: x =  - 2 \: or  - 3}}}

❒ p(x) = x^2 + 2x + 1 = 0

\large{ \rm{ \longrightarrow \: x =  \dfrac{  - 2 \pm  \sqrt{ {2}^{2}  - 4 \times 1 \times 1} }{2 \times 1} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - 2 \pm \sqrt{4 - 4} }{2} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - 2 \pm 0}{2} }}

So here,

\large{\boxed{ \rm{ \longrightarrow \: x =  - 1 \: or \:  - 1}}}

❒ p(x) = x^2 - x - 20 = 0

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - ( - 1) \pm  \sqrt{( - 1) {}^{2} - 4 \times 1 \times ( - 20) } }{2 \times 1} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{ 1 \pm \sqrt{1 + 80} }{2} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{1 \pm 9}{2} }}

So here,

\large{\boxed{ \rm{ \longrightarrow \: x = 5 \: or \:  - 4}}}

❒ p(x) = x^2 - 3x - 4 = 0

\large{ \rm{ \longrightarrow \: x =   \dfrac{  - ( - 3) \pm \sqrt{( - 3) {}^{2} - 4 \times 1 \times ( - 4) } }{2 \times 1} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{3 \pm \sqrt{9  + 16} }{2 \times 1} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{3  \pm 5}{2} }}

So here,

\large{\boxed{ \rm{ \longrightarrow \: x = 4 \: or \:  - 1}}}

<u>━━━━━━━━━━━━━━━━━━━━</u>

5 0
2 years ago
Read 2 more answers
Shen is saving money to buy a game. The game costs $24 , and so far he has saved two-thirds of this cost. How much money has She
valentinak56 [21]
24 divided by 3 = 8

but thers two third so 8+8=16

shen has saved $16.00

Hope this helps!
8 0
3 years ago
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