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nlexa [21]
3 years ago
12

An arts academy required there to be 4 teachers for every 64 students and 6 tutors for every 48 students.How many students does

the academy have per teacher? Per tutor? How many teachers does the academy need if it has 96 students?
Mathematics
1 answer:
irina [24]3 years ago
3 0

Answer:

1. 16 teachers  2. 8 tutor  3. 6 teachers

Step-by-step explanation:

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What is the missing fact in this fact family 7+5 =12 , 12-5=7 , 12-7=5
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Missing fact is 13 seats are empty
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4 years ago
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The mean length of 7 planks of wood is 1.35m. When an extra plank of wood is added, the mean length of a plank of wood increase
azamat

Answer: Hello There!................

Mean = Sum of observation / total number of observation

Mean = 1.35m

1.35  = Sum of length of 7 wood planks/ 7

1.35*7 = Sum of the length of 7 wood planks

9.45 = Sum of the length of 7 wood planks

Let the length of extra added wood plank = x

So,

(Sum of the length of 7 wood planks + x) / 8 = 1.4m

(9.45 + x) / 8 = 1.4

9.45 + x = 1.4 * 8

9.45 + x = 11.2

x = 11.2- 9.45

= 1.75

So,

The length of extra added wood plank = 1.75m

Step-by-step explanation:

Mark me brainest please. Hope this helps. Anna ♥

6 0
2 years ago
Solve the following system of equations. <br> 2x + y = 3 <br> x = 2y - 1
Shalnov [3]
2(2y-1)+y=3
y=1
x=2 x1 - 1
x=1
7 0
3 years ago
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Solve the quadratic equation 6x2 + 1 = 5x. Show your work.
Artyom0805 [142]

Answer:

<h2>Factors = (3x - 1) (2x - 1)</h2><h2>values :- x = 1/3 , 1/2</h2>

Step-by-step explanation:

= > 6x^2 + 1 = 5x

• Bring it in the standard form,

= > 6x^2 - 5x + 1 = 0

= > 6x^2 - (3 + 2)x + 1 = 0

= > 6x^2 - 3x - 2x + 1 = 0

• Take out common

= > 3x (2x - 1) - 1 (2x - 1) = 0

= > (3x - 1) (2x - 1) = 0...factors

= > x = 1/3 and 1/2... values of x

<h2>Hope it helps you!! </h2>

7 0
2 years ago
Unsure how to do this calculus, the book isn't explaining it well. Thanks
krok68 [10]

One way to capture the domain of integration is with the set

D = \left\{(x,y) \mid 0 \le x \le 1 \text{ and } -x \le y \le 0\right\}

Then we can write the double integral as the iterated integral

\displaystyle \iint_D \cos(y+x) \, dA = \int_0^1 \int_{-x}^0 \cos(y+x) \, dy \, dx

Compute the integral with respect to y.

\displaystyle \int_{-x}^0 \cos(y+x) \, dy = \sin(y+x)\bigg|_{y=-x}^{y=0} = \sin(0+x) - \sin(-x+x) = \sin(x)

Compute the remaining integral.

\displaystyle \int_0^1 \sin(x) \, dx = -\cos(x) \bigg|_{x=0}^{x=1} = -\cos(1) + \cos(0) = \boxed{1 - \cos(1)}

We could also swap the order of integration variables by writing

D = \left\{(x,y) \mid -1 \le y \le 0 \text{ and } -y \le x \le 1\right\}

and

\displaystyle \iint_D \cos(y+x) \, dA = \int_{-1}^0 \int_{-y}^1 \cos(y+x) \, dx\, dy

and this would have led to the same result.

\displaystyle \int_{-y}^1 \cos(y+x) \, dx = \sin(y+x)\bigg|_{x=-y}^{x=1} = \sin(y+1) - \sin(y-y) = \sin(y+1)

\displaystyle \int_{-1}^0 \sin(y+1) \, dy = -\cos(y+1)\bigg|_{y=-1}^{y=0} = -\cos(0+1) + \cos(-1+1) = 1 - \cos(1)

7 0
1 year ago
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