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bixtya [17]
3 years ago
10

you are a school for tographer taking individual and class pictures for two classes of 21 students each on average each individu

al picture take 3 minutes and a class picture take 10 minutes about how long should it take you to take all of the pictures
Mathematics
1 answer:
sveta [45]3 years ago
3 0
It would take 2 hours and 26 minutes
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Simplify: -0.8b+4.1c-(-3.2b)-0.1c
BigorU [14]

[ Answer ]

\boxed{2.4b \ + \ 4c}

[ Explanation ]

  • Simplify: -0.8b + 4.1c - (-3.2b) - 0.1c

-----------------------------------

  • Apply Rule: - (-a) = a

-0.8b + 4.1c + 3.2b - 0.1c

  • Group Like Terms

-0.8b + 3.2b + 4.1c - 0.1c

  • Add Similar Elements: -0.8b + 3.2b = 2.4b

2.4b + 4.1c - 0.1c

  • Add Similar Elements: 4.1c - 0.1c = 4c

2.4b + 4c

\boxed{[ \ Eclipsed \ ]}

8 0
3 years ago
What is the quotient (125 – 8x3) ÷ (25 + 10x + 4x2)?
ahrayia [7]
Your answer is:
(125 - 8x^3) / (25 + 10x + 4x^2) = - ((2x - 5) * (4x^2 + 10x + 25)) / (4x^2 + 10x + 25) = - (2x - 5) = - 2x + 5

The correct result would be - 2x + 5.
6 0
3 years ago
Read 2 more answers
Trip has 15 coins worth 95 cents.Four of the coins are each worth twice as much as the rest. Construct a math argument to justif
const2013 [10]
(11*5)+(4*10)=95
11+4=15
Both are equal to the constraints of the word problem and there for justified.
6 0
3 years ago
(WILL GIVE BRINALIST TO BEST ANWER) WHATS the measure of < 3
malfutka [58]

Answer:

∠3 = 60°

Step-by-step explanation:

Since g and h are parallel lines then

∠1 and ∠2 are same side interior angles and are supplementary, hence

4x + 36 +3x - 3 = 180

7x + 33 = 180 ( subtract 33 from both sides )

7x = 147 ( divide both sides by 7 )

x = 21

Thus ∠2 = (3 × 21) - 3 = 63 - 3 = 60°

∠ 2 and ∠3 are alternate angles and congruent, hence

∠3 = 60°

8 0
2 years ago
What is the quadratic regression equation for the data set? ​ ​ ​ <br><br> PLEASEE HELP ME
Soloha48 [4]
You're trying to find constants a_0,a_1,a_2 such that \hat y=a_0+a_1\hat x+a_2{\hat x}^2. Equivalently, you're looking for the least-square solution to the following matrix equation.


\underbrace{\begin{bmatrix}1&6&6^2\\1&3&3^2\\\vdots&\vdots&\vdots\\1&9&9^2\end{bmatrix}}_{\mathbf A}\underbrace{\begin{bmatrix}a_0\\a_1\\a_2\end{bmatrix}}_{\mathbf x}=\underbrace{\begin{bmatrix}100\\110\\\vdots\\70\end{bmatrix}}_{\mathbf b}

To solve \mathbf{Ax}=\mathbf b, multiply both sides by the transpose of \mathbf A, which introduces an invertible square matrix on the LHS.

\mathbf{Ax}=\mathbf b\implies\mathbf A^\top\mathbf{Ax}=\mathbf A^\top\mathbf b\implies\mathbf x=(\mathbf A^\top\mathbf A)^{-1}\mathbf A^\top\mathbf b

Computing this, you'd find that

\mathbf x=\begin{bmatrix}a_0\\a_1\\a_2\end{bmatrix}\approx\begin{bmatrix}121.119\\-3.786\\-0.175\end{bmatrix}

which means the first choice is correct.
8 0
3 years ago
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