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iren2701 [21]
4 years ago
9

Which statement justifies whether or not the two quadrilaterals are congurent

Mathematics
1 answer:
aksik [14]4 years ago
8 0

Answer:

is there photo

Step-by-step explanation:

please post the photo

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Solving a Multi-Step Equation
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x = 15

Step-by-step explanation:

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I have a fair six sided dice numbered 4,9,12,16,20 and 24 what is the probability of rolling a multiple of 4
steposvetlana [31]
It would be 5/6 because 4 can go into 4,12,16,20, and 24.

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In 0.724, which digit is in the thousandths place?
ikadub [295]

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4

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6 0
3 years ago
Which description matches the graph of the inequality y ≥ |x + 2| – 3? a shaded region above a solid boundary line a shaded regi
Alex777 [14]
The given inequality is y ≥ |x + 2| -3.

This inequality may be written two ways:
(a) y ≥ x + 2 - 3
    or
    y ≥ x - 1

(b) y ≥ -x -2 - 3
    or
     y ≥ -x - 5

A graph of the inequality is shown below. The shaded region satisfies the inequality.

Answer: A shaded region above a solid boundary line.

6 0
3 years ago
A man rides a bicycle a distance of 48 miles in a certain amount of time. If the man increased his speed by 2 miles per hour, he
seropon [69]
Recall your d = rt, distance = rate * time.

so, in his usual speed, say is hmm "r" mph, so in his usual "r" speed, he rolls on for the 48 miles and it takes him "t" hours to get there.

now, if he increases his speed by 2 mph, then his new speed is "r + 2", and he arrives there 4 hours earlier, so if he took "t" hours going at "r" speed, then when he's going faster at "r + 2", he only takes "t - 4" hours, for the same 48 miles.

\bf \begin{array}{lccclll}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
\textit{usual speed}&48&r&t\\
\textit{faster speed}&48&r+2&t-4
\end{array}
\\\\\\
\begin{cases}
48=rt\implies \frac{48}{r}=\boxed{t}\\
48=(r+2)(t-4)\\
----------\\
48=(r+2)\left( \boxed{\frac{48}{r}}-4 \right)
\end{cases}
\\\\\\
48=(r+2)\left( \cfrac{48-4r}{r} \right)\implies 48r=(r+2)(48-4r)


\bf 48r=48r-4r^2+96-8r\implies 0=-4r^2+96-8r
\\\\\\
4r^2+8r-96=0\implies r^2+2r-24=0
\\\\\\
(r+6)(r-4)=0\implies r=
\begin{cases}
-6\\
\boxed{4}
\end{cases}

it cannot be a negative value, since is just a forward speed rate, so can't be -6.
8 0
3 years ago
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