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kramer
3 years ago
6

HELP

Mathematics
1 answer:
madreJ [45]3 years ago
8 0
It’s to hard to see what you need help on e
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There are 14 teams in a basketball league.
Talja [164]

Answer:

Step-by-step explanation:

Given that :

Number of teams = 14

Each team plays every other team twice ;

Using the combination formula :

nCr = n! ÷ (n-r)! r!

14C2 = 14! ÷ (14 - 2)! 2!

14C2 = 14! ÷ (12)! 2!

14C2 = (14 * 13) ÷ 2 * 1

14C2 = 182 / 2

14C2 = 91

Hence, since they are going to be playing each other twice :

2(14C2)

2 * 91 = 182games

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3 years ago
The square of the difference of a number and 3
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(x-3)^2 x=a number in this case
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Which of the following processes requires the production of gametes
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The answer would be B) Sexual Reproduction
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Of a total of 6,000.00, part was invested at a simple interest rate of 8% and part at 7% if the total interest income for one ye
alekssr [168]

Answer: Ok so first 0.05(6000 - x) + 0.07x = 372

300 - 0.05x + 0.07x = 372

300 + 0.02x = 372

0.02x = 72

x = 3600

He invested $3600 at 7% and $2400 at 5%.

<h2>Hope this helps have a bless day❤️</h2>

Step-by-step explanation:

6 0
3 years ago
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Calculate the area of the triangle with the following vertices (3, -7), (6, 4), (-2, -3)
Monica [59]

Answer:

\boxed{\mathsf{A} \triangle = \red{\dfrac{67}{2}u.a}}

Step-by-step explanation:

Let's follow up with the solution. Considering a triangle with the vertices \mathsf{A(x_A, y_A)}, \mathsf{B(x_B, y_B)} and \mathsf{C(x_C, y_C)}, have a look at the representation in the cartesian plan.

From this representation we can say that the area (A) of a triangle through the knowledge of <u>analytical geometry</u> is given by the determinant of the vertices divided by two, mathematically,

\mathsf{A} \triangle =  \dfrac{\left| \begin{array}{ccc}  \mathsf{x_A} & \mathsf{y_A }& 1 \\  \mathsf{x_B} &  \mathsf{ y_B} & 1 \\ \mathsf{ x_C} &  \mathsf{ y_C} & 1 \end{array} \right|}{2}

So, applying this knowledge we're going to have,

\mathsf{A} \triangle =  \dfrac{\left| \begin{array}{ccc}  3 & -7 & 1 \\ 6 &  4 & 1 \\ -2 &  -3 & 1 \end{array} \right|}{2}

\mathsf{A} \triangle =  \dfrac{1}{2}\left[  \left.\begin{array}{ccc}   3 & -7 & 1 \\ 6 &  4 & 1 \\ -2&  -3 & 1 \end{array}  \right| \begin{array}{cc} 3 & -7 \\ 6 & 4 \\ -2 & -3 \end{array} \right]

\mathsf{A} \triangle = \dfrac{12 + 14 - 18 - (-8 - 9 - 42)}{2}

\red{\mathsf{A} \triangle = \dfrac{67}{2} = 33,5u.a}

Hope you enjoy it, see ya!)

\green{\mathsf{FROM}}: Mozambique, Maputo – Matola City – T-3

DavidJunior17

3 0
3 years ago
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