Please answer all part correct

Mathematics

1 answer:

insens350 [35]3 years ago

7 0

Answer:

a). We want to know how much each point was worth.

b). $12x+5=41$

c). Each problem worth 3 points.

Step-by-step explanation:

a). We want to know how much each problem was worth. Because we have the total points of the test, and how much was the bonus. But we still don't know the worth of each problem.

b). We know that the total points of the test were 41, and the bonus 5 points. There were 12 problems on the test and we are going to use "x" for the unknown part (how many points each problem was worth).

The equation is :

$12x+5=41$

Why 12x? Because if you multiply the twelve problems of the test with the worth of each one and then add the 5 points of the bonus you will obtain the total points of the test (41).

c). Now we have to solve the equation, this means that we have to clear "x":

$12x+5=41$

Subtract 5 from both sides.

$12x+5-5=41-5\\12x=36$

Finally divide in 12 both sides of the equation:

$\frac{12x}{12}=\frac{36}{12} \\\\x=3$

Then each problem worth 3 points.

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URGENT WILL GIVE BRAINLIEST solve using matrices 4x + 5y = 40 x – y = 1

dalvyx [7]

Answer:

x = 5, y = 4

Step-by-step explanation:

$\left\{\begin{array}{ccc}4x+5y=40\\x-y=1\end{array}\right\\\\A=\left[\begin{array}{ccc}4&5\\1&-1\end{array}\right]\\\\detA=\left|\begin{array}{ccc}4&5\\1&-1\end{array}\right|=(4)(-1)-(1)(5)=-4-5=-9\\\\A^D=\left[\begin{array}{ccc}-1&-1\\-5&4\end{array}\right]\\\\\left(A^D\right)^T=\left[\begin{array}{ccc}-1&-5\\-1&4\end{array}\right]$

$A^{-1}=\dfrac{1}{detA}\left(A^D\right)^T\\\\A^{-1}=\dfrac{1}{-9}\left[\begin{array}{ccc}-1&-5\\-1&4\end{array}\right]=\left[\begin{array}{ccc}\dfrac{1}{9}&\dfrac{5}{9}\\\dfrac{1}{9}&-\dfrac{4}{9}\end{array}\right]$

$A\cdot\left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}40\\1\end{array}\right] \\\\A^{-1}A\cdot\left[\begin{array}{ccc}x\\y\end{array}\right] =A^{-1}\cdot\left[\begin{array}{ccc}40\\1\end{array}\right] \\\\\left[\begin{array}{ccc}x\\y\end{array}\right] =A^{-1}\cdot\left[\begin{array}{ccc}40\\1\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}\dfrac{1}{9}&\dfrac{5}{9}\\\dfrac{1}{9}&-\dfrac{4}{9}\end{array}\right]\cdot\left[\begin{array}{ccc}40\\1\end{array}\right] \\\\\left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}\left(\frac{1}{9}\right)(40)+\left(\frac{5}{9}\right)(1)\\\\\left(\frac{1}{9}\right)(40)+\left(-\frac{4}{9}\right)(1)\end{array}\right]\\\\\left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}5\\4\end{array}\right]\Rightarrow x=5,\ y=4$