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Ira Lisetskai [31]
3 years ago
10

A. 18 B. 8 C. 8.5 D. 9

Mathematics
1 answer:
Karolina [17]3 years ago
8 0

A = l x t

153 = 17x

x = 9

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Basile [38]
Your answer should be, 3x= 22y to the tenth power...
6 0
3 years ago
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Novay_Z [31]
Try this:
1) if ratio=100/3, then 100/3=2600/x
x=26*3=78 p.
2) if (-2) ⇒ -3*(4)=-12 ⇒ answer: (-2;-12)
if (-1) ⇒ -3*(1)=-3 ⇒ answer: (-1;-3)
if 0 ⇒ -3*0=0 ⇒ answer: (0;0)
if 1 ⇒ -3*1=-3 ⇒ answer (1;-3)
if 2 ⇒ -3*4=-12 ⇒ answer: (2;-12)
7 0
4 years ago
Please help me!<br> simplify. 5-[-(-20] <br> a. -3<br> b.3<br> c.-7<br> d.7
frosja888 [35]

Im sorry i dont know

8 0
4 years ago
Question 101 points)
maw [93]

Answer:

\large\boxed{y=-\dfrac{2}{3}x+\dfrac{13}{3}}

Step-by-step explanation:

\text{The slope-intercept form of an equation of a line:}\\\\y=mx+b\\\\m-slope\\b-y-intercept\\\\\text{The formula of a slope:}\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\\\\===============================

\text{We have the point:}\\\\(5,\ 1)\ \text{and}\ (-4,\ 7).\ \text{Substitute:}\\\\m=\dfrac{7-1}{-4-5}=\dfrac{6}{-9}=-\dfrac{6:3}{9:3}=-\dfrac{2}{3}\\\\\text{We have the equation in form:}\\\\y=-\dfrac{2}{3}x+b\\\\\text{Put the coordinates of the point (5, 1) to the equation:}\\\\1=-\dfrac{2}{3}(5)+b\\\\1=-\dfrac{10}{3}+b\qquad\text{add}\ \dfrac{10}{3}\ \text{to the both sides}\\\\\dfrac{3}{3}+\dfrac{10}{3}=b\to b=\dfrac{13}{3}\\\\\text{Finally:}\\\\y=-\dfrac{2}{3}x+\dfrac{13}{3}

4 0
3 years ago
The numbers of teams remaining in each round of a single-elimination tennis tournament represent a geometric sequence where an i
Anit [1.1K]

Answer:

a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}

Step-by-step explanation:

We are given the following in the question:

The numbers of teams remaining in each round follows a geometric sequence.

Let a be the first the of the geometric sequence and r be the common ration.

The n^{th} term of geometric sequence is given by:

a_n = ar^{n-1}

a_4 = 16 = ar^3\\a_6 = 4 = ar^5

Dividing the two equations, we get,

\dfrac{16}{4} = \dfrac{ar^3}{ar^5}\\\\4}=\dfrac{1}{r^2}\\\\\Rightarrow r^2 = \dfrac{1}{4}\\\Rightarrow r = \dfrac{1}{2}

the first term can be calculated as:

16=a(\dfrac{1}{2})^3\\\\a = 16\times 6\\a = 128

Thus, the required geometric sequence is

a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}

4 0
3 years ago
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