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11Alexandr11 [23.1K]
3 years ago
5

The product of 9 and t squared, increased by the sum of the square of t anne 2

Mathematics
1 answer:
Serjik [45]3 years ago
7 0

Answer:

10t2+ 2

Step-by-step explanation:

9t^{2} x (t^{2} + 2) add 9t2, and t2 to get 10t2 + 2

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I need help with this question. It is a file.
olchik [2.2K]
D because coefficient is the number next to the variable and 7 is the coefficient to the fifth power
7 0
3 years ago
Will enyone help me answer this question please
SSSSS [86.1K]
You could use 4(division) sign 25 + 3 (division sign) 15 * 10, ok?
6 0
4 years ago
In a particular region, for families with a combined income of $75,000 or more, 15% of these families have no children, 35% of t
SCORPION-xisa [38]

Answer:

The probability distribution for x:"number of children per family for this income group" is:

\text{P(x=0)}=0.15\\\\\text{P(x=1)}=0.35\\\\\text{P(x=2)}=0.45\\\\\text{P(x=3)}=0.05\\\\

Step-by-step explanation:

With the information given we have the relative frequencies of each category.

We know:

\text{P(x=0)}=0.15\\\\\text{P(x=1)}=0.35\\\\\text{P(x=2)}=0.45\\\\\text{P(x=3)}=0.05\\\\

7 0
3 years ago
Find the equation of the straight line with the gradient -2/3 and passing through the point(1,1)
adell [148]

Answer:

y = -2/3 x + 5/3

Step-by-step explanation:

y = mx + b

y = -2/3 x + b

1 = -2/3 × 1 + b

3/3 + 2/3 = b

b = 5/3

y = -2/3 x + 5/3

4 0
2 years ago
Find the equation of the parabola with focus (5, 1) and directrix y = -1.
Mumz [18]
Check the picture below.

since the focus point is above the directrix, that simply means the parabola is a vertical one, and therefore the square variable is the "x".

keeping in mind that, there's a distance "p" from the vertex to either the focus point or the directrix, that puts the vertex half-way between those fellows, in this case at 0, right between 1 and -1, as you see in the picture, and the parabola looks like so.

since the parabola is opening upwards, the value for "p" is positive, thus

\bf \textit{parabola vertex form with focus point distance}
\\\\
\begin{array}{llll}
4p(x- h)=(y- k)^2
\\\\
\boxed{4p(y- k)=(x- h)^2}
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ( h, k)\\\\
 p=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------\\\\
\begin{cases}
h=5\\
k=0\\
p=1
\end{cases}\implies 4(1)(y-0)=(x-5)^2
\\\\\\
4y=(x-5)^2\implies  y=\cfrac{1}{4}(x-5)^2

8 0
3 years ago
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