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IgorC [24]
3 years ago
8

Translate to an​ equation, then solve

Mathematics
1 answer:
rjkz [21]3 years ago
4 0

Answer:

n=12

Step-by-step explanation:

13 less than 9 times a number: 13 less than 9n, 9n-13

11 more than 7 times a number: 11 more than 7n, 7n+11

9n-13=7n+11

2n=24

n=12

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Haos dog ate 40 cns of dog food in 31 day. how many cans should hao buy to feed his dog in 6 days
Crank
The answer is 8. 40 / 31 = 1.29032258. Multiply that by 6 and you have 7.74193548. Round up because you can't buy 7.7 cans of dog food.
7 0
3 years ago
How to I find the 3rd derivative of a function
Margaret [11]
By simplifing i think

3 0
3 years ago
Need help with this :(
emmainna [20.7K]

Answer:

B.    m = \frac{3}{2}, y-int = 3

Step-by-step explanation:

To find the slope of the line, we need to plot two points

(2, 6) (-2, 0)

m = slope

Then use the Slope-Formula to Solve for the slope

m = \frac{y2-y1}{x2-x1}        

m = \frac{0-6}{-2-2}

m = \frac{-6}{-4}

m = \frac{6}{4}

m = \frac{3}{2}

The Slope is \frac{3}{2}

The y-intercept is 3 because the line is going 3 units upwards on y axis!

Equation: y = 3/2x + 3

Slope: 3/2

y-intercept: 3

5 0
3 years ago
Read 2 more answers
Check answer please
Cerrena [4.2K]
The fourth or the D) Option is correct.

To find the new induced matrix via a scalar quantified multiplication we have to multiply the scalar quantity with each element surrounded and provided in a composed (In this case) 3×3 or three times three matrix comprising 3 columns and 3 rows for each element which is having a valued numerical in each and every position.

Multiply the scalar quantity with each element with respect to its row and column positioning that is,

Row × Column. So;

(1 × 1) × 7, (2 × 1) × 7, (3 × 1) × 7, (1 × 2) × 7, (2 × 2) × 7, (3 × 2) × 7, (1 × 3) × 7, (2 × 3) × 7 and (3 × 3) × 7. This will provide the final answer, that is, the D) Option.

To interpret and make it more interesting in LaTeX form. Here is the solution with LaTeX induced matrix.

\mathcal{A = \begin{bmatrix}1 & 0 & 3 \\ 2 & -1 & 2 \\ 0 & 2 & 1 \\ \end{bmatrix}}

\mathbf{\therefore \quad 7A = 7 \times \begin{bmatrix}1 & 0 & 3 \\ 2 & - 1 & 2 \\ 0 & 2 & 1 \\ \end{bmatrix}}

\mathbf{\therefore \quad \begin{bmatrix}7 \times 1 & 7 \times 0 & 7 \times 3 \\ 7 \times 2 & 7 \times -1 & 7 \times 2 \\ 7 \times 0 & 7 \times 2 & 7 \times 1 \\ \end{bmatrix}}

\therefore \quad \begin{\bmatrix}7 & 14 & 0 \\ 0 & -7 & 14 \\ 21 & 14 & 7 \end{bmatrix}

Hope it helps.
5 0
3 years ago
Part of a tiling design is shown. The center is a regular hexagon. A square is on each side of the hexagon, and an equilateral t
LiRa [457]
Area of the squares (6):
A 1 = 6 · 10² = 600 in²
Area of the equilateral triangles (6):
A 2 = 6 · 10²√3 / 4 = = 6 · 25 · 1.73 = 259.8 in²
Area of the hexagon:
A 3 = 6 · 10²√3/4 = 259.8 in²
Total area:
600 + 259.8 + 259.8 = 1,119.6 in²
Answer: C ) 
3 0
3 years ago
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