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Ierofanga [76]
2 years ago
12

Three dice, each with faces numbered 1 through 6, are arranged as shown. Seven faces are visible. Find the sum of the numbers on

all the faces that are not visible.

Mathematics
2 answers:
In-s [12.5K]2 years ago
6 0
The Answer for your math problem is 37
PolarNik [594]2 years ago
3 0

Answer:

37

Step-by-step explanation:

First, we need to find the sum of all the numbers of the dice. The expression 3(1 + 2 + 3 + 4 + 5 + 6), or 3(21) does that for us. 3 × 21 = 63. So the sum of all of the numbers of the dice is 63. But we already know 7 of the numbers there: 2, 3, 1, 6, 5, 4, 5. In the expression 3(21), we counted these numbers as well. So now we can add these 7 numbers together then subtract them from 3(21), or 63. We can write the expression 3(21) - (2 + 3 + 1 + 6 + 5 + 4 + 5), or 3(21) - 26 = 63 - 26 = 37.

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Find f(x) = -3x2 + 2x - 7 if f(-1) <br> I am really confused. Can someone please help me?
Oksanka [162]
F(-1)
replace all x with -1
f(-1)=-3(-1)^2+2(-1)-7
f(-1)=-3(1)-2-7
f(-1)=-3-9
f(-1)=-12
3 0
4 years ago
4z-(-3z) can you help me​
Korvikt [17]

Answer:

7z

Step-by-step explanation:

4z-(-3z).

Two negatives equal a positive only <u>when they are right next to each other.</u>

4z+3z=7z. Hope this helps :D

8 0
3 years ago
HELP!! ANSWER AND SHOW STEPS IF POSSIBLE! ASAP HELP PLEASE
Schach [20]
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APR = 18.6%, thus monthly interest rate = 18.6 / 12 = 1.55%
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2.) Previous balance = 5834.53
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5 0
4 years ago
Read 2 more answers
The vertex of this parabola is at (2,-1) when the y value is 0 and then x value is 5 what is the coefficient of the squared term
Darina [25.2K]

\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \stackrel{\textit{we'll use this one}}{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{2}{ h},\stackrel{-1}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=2\\ k=-1 \end{cases}\implies y=a(x-2)^2-1 \\\\\\ \textit{we also know that } \begin{cases} y=0\\ x=5 \end{cases}\implies 0=a(5-2)^2-1\implies 1=9a \\\\\\ \cfrac{1}{9}=a\qquad therefore\qquad \boxed{y=\cfrac{1}{9}(x-2)^2-1}


now, let's expand the squared term to get the standard form of the quadratic.


\bf y=\cfrac{1}{9}(x-2)^2-1\implies y=\cfrac{1}{9}(x^2-4x+4)-1 \\\\\\ y=\cfrac{1}{9}x^2-\cfrac{4}{9}x+\cfrac{4}{9}-1\implies \stackrel{its~coefficient}{y=\stackrel{\downarrow }{\cfrac{1}{9}}x^2-\cfrac{4}{9}x-\cfrac{5}{9}}

4 0
3 years ago
Read 2 more answers
Explain answer !! <br><br> ( WILL GIVE BRAINLEST)
OLEGan [10]

Larissa: y = 250 - 25x

Chucho: y = 30 +30x

4 weeks

Step-by-step explanation: It says it in the problem, and you can see they cross at 4 weeks

4 0
3 years ago
Read 2 more answers
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