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marissa [1.9K]
3 years ago
8

HELP ME PLEASE ❗❗❗❗

Mathematics
1 answer:
Archy [21]3 years ago
3 0

Answer:

Q3 = 30

Step-by-step explanation:

minimum value = 5

Q1 = 20  (this is the median of the lower half of data)

Median (sometimes called Q2) = 25

Q3 = 30 (this is the median of the upper half of data)

maximum value = 45

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What is the polynomial -9+1/8x-7x^4+x^2-x^5 in standard form
Oduvanchick [21]

Answer:

-x^5 - 7x^4 + x^2 + (1/8)x - 9

Step-by-step explanation:

Rewrite the given -9+1/8x-7x^4+x^2-x^5 in descending order of the variable x:

-9+1/8x

-x^5 - 7x^4 + x^2 + (1/8)x - 9

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The slope intercept form is y=mx+b, and b represents the y-intercept. In the given equation C=45t+100, 100 is the y-intercept. It represents the one-time fee to join the gym membership.

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The coordinates of rhombus ABCD are A(–4, –2), B(–2, 6), C(6, 8), and D(4, 0). What is the area of the rhombus? Round to the nea
a_sh-v [17]
Check the picture below.

so the rhombus has the diagonals of AC and BD, now keeping in mind that the diagonals bisect each, namely they cut each other in two equal halves, let's find the length of each.

\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
A(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-2})\qquad 
C(\stackrel{x_2}{6}~,~\stackrel{y_2}{8})\qquad \qquad 
%  distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{[6-(-4)]^2+[8-(-2)]^2}\implies AC=\sqrt{(6+4)^2+(8+2)^2}
\\\\\\
AC=\sqrt{10^2+10^2}\implies AC=\sqrt{10^2(2)}\implies \boxed{AC=10\sqrt{2}}\\\\
-------------------------------

\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
B(\stackrel{x_1}{-2}~,~\stackrel{y_1}{6})\qquad 
D(\stackrel{x_2}{4}~,~\stackrel{y_2}{0})\qquad \qquad BD=\sqrt{[4-(-2)]^2+[0-6]^2}
\\\\\\
BD=\sqrt{(4+2)^2+(-6)^2}\implies BD=\sqrt{6^2+6^2}
\\\\\\
BD=\sqrt{6^2(2)}\implies \boxed{BD=6\sqrt{2}}

that simply means that each triangle has a side that is half of 10√2 and another side that's half of 6√2.

namely, each triangle has a "base" of 3√2, and a "height" of 5√2, keeping in mind that all triangles are congruent, then their area is,

\bf \stackrel{\textit{area of the four congruent triangles}}{4\left[ \cfrac{1}{2}(3\sqrt{2})(5\sqrt{2}) \right]\implies 4\left[ \cfrac{1}{2}(15\cdot (\sqrt{2})^2) \right]}\implies 4\left[ \cfrac{1}{2}(15\cdot 2) \right]
\\\\\\
4[15]\implies 60

7 0
3 years ago
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Neko [114]

Answer:

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Step-by-step explanation:

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5 0
3 years ago
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Nitella [24]
The answer to this question is -3/35
3 0
3 years ago
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