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Setler [38]
3 years ago
10

A box of 42 chocolates contains milk and dark chocolates in the ratio of 3:4. How many chocolates of each type there are?

Mathematics
1 answer:
Elza [17]3 years ago
4 0
<h2>Answer:The number of milk chocolates are 18 and the number of dark chocolates are 24.</h2>

Step-by-step explanation:

Let the number of milk chocolates be m.

Let the number of dark chocolates be d.

Given that the box contains the milk and dark chocolates in the ratio 3:4.

So,\frac{m}{d}=\frac{3}{4}               ...(i)

Given that the total number of chocolates are 42.

So,m+d=42.                                   ...(ii)

Using (i) and (ii),

m+\frac{4}{3}m=42

\frac{7}{3}m=42

m=18

d=\frac{4}{3}m=\frac{4}{3}\times 18=24

So,the number of milk chocolates are 18 and the number of dark chocolates are 24.

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4 0
2 years ago
A student finds the sum of the angle measures in an octagon by multiplying 7 • 180°. What is the student's error?​
Lostsunrise [7]

The student should compute 6*180 = 1080 because of the formula

S = 180(n-2)

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7 0
3 years ago
48
Diano4ka-milaya [45]
2.20 kg + 1.90 kg

*1*
2.20
+1.90
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4 0
3 years ago
Help please!!!!!!!!!
In-s [12.5K]

ANSWER

24


EXPLANATION

For a matrix A of order n×n, the cofactor C_{ij} of element a_{ij} is defined to be


   C_{ij} = (-1)^{i+j} M_{ij}


M_{ij} is the minor of element a_{ij} equal to the determinant of the matrix we get by taking matrix A and deleting row i and column j.


Here, we have


   C_{11} = (-1)^{1+1} M_{11} = M_{11}


M₁₁ is the determinant of the matrix that is matrix A with row 1 and column 1 removed. The bold entries are the row and the column we delete.


   \begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right)  \end{aligned}


Since the determinant of a 2×2 matrix is


   \det\left(  \begin{bmatrix} a & b \\ c& d  \end{bmatrix} \right) = ad-bc


it follows that


   \begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right) \\ &= (0)(-8) - (-3)(8) \\ &= -(-24) \\ &= 24 \end{aligned}


so C_{11} = M_{11} = 24

4 0
3 years ago
Look at attachment 10 POINTS!!!
Leni [432]

Answer:

The midpoint of TS is (-1,-3)

The coordinates of M should be (8,18)

Step-by-step explanation:


3 0
4 years ago
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