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Katen [24]
3 years ago
15

What is the standard form of -1

Mathematics
2 answers:
Irina18 [472]3 years ago
7 0
Negative one (the standard form is a number written out like 25 is twenty five)
laiz [17]3 years ago
5 0

Answer:

Negative one

Step-by-step explanation:

Standard form is when you write out the number.

ex. 45 is forty-five

     102 is one hundred two

Hope this helps!

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Write the equation of a circle with center (6,4) that passes through the coordinate (2,1). In your final answer, include all of
kari74 [83]
R^2=(x-6)^2+(y-4)^2

r^2=(6-2)^2+(4-1)^2, r^2=16+9=25

(x-6)^2+(y-4)^2=25
8 0
3 years ago
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Sequence of number - 25,23,20,16<br>do with process​
Nitella [24]

Answer:

what should we be finding?

8 0
3 years ago
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Need help with this. Please
kkurt [141]

Answer:

46

Step-by-step explanation:

6 times 2 is 12

23 times 2 is 46

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3 years ago
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Which of the following describes the probability distribution below?
7nadin3 [17]

Answer:

Answer: A.) The mean is greater than the median, and the majority of the data points are to the left of the mean.

It is clear that most of the data (around 75%) is consist of value 1, which is the leftmost part of the data. Since it was more than 50% of the data, the median should be 1.

if 75% data is 1, it need 25% data with value at least 5 to make the means equal to 2. The means would be bigger than 1 but less than 2, so most(75% data is 1) of the data would be on the left of the mean.

8 0
2 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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