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3 years ago

Not good at this need some help

Mathematics

1 answer:

emmasim [6.3K]3 years ago

6 0

Answer:

216 sq. units

Step-by-step explanation:

From the figure, we can see that only one pair of opposite angles are equal. So, the quadrilateral is a kite.

Formula to find the area of a kite:

Area, A = $\frac{1}{2} \times d_1 \times d_2$

where, $d_1$ and $d_2$ are the lengths of the diagonals.

Here, $d_1 = 18$ units.

And, $d_2 = 24$ units.

Therefore, the area A = $\frac{1}{2} \times 18 \times 24$

= $\frac{432}{2}$

= 216 sq. units which is the required answer.

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2 years ago

Given x^2 -4x+4 is a factor of the polynomial 2x^4 - 11x^3 +68x-80, completely factor the polynomial using algebraic methods.

asambeis [7]

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2 years ago

Please help me with the below question.

tresset_1 [31]

We have the following three conclusions about the <em>piecewise</em> function evaluated at x = 14.75:

$\lim_{t \to 14.75^{-}} f(t) = 66$ .
$\lim_{t \to 14.75^{+}} f(t) = 10$ .
$\lim_{t \to 14.75} f(t)$ does not exist as $\lim_{t \to 14.75^{-}} f(t) \ne \lim_{t \to 14.75^{+}} f(t)$ .

<h3>How to determinate the limit in a piecewise function</h3>

In a <em>piecewise</em> function, the limit for a given value exists when the two <em>lateral</em> limits are the same and, thus, continuity is guaranteed. Otherwise, the limit does not exist.

According to the definition of <em>lateral</em> limit and by observing carefully the figure, we have the following conclusions:

$\lim_{t \to 14.75^{-}} f(t) = 66$ .
$\lim_{t \to 14.75^{+}} f(t) = 10$ .
$\lim_{t \to 14.75} f(t)$ does not exist as $\lim_{t \to 14.75^{-}} f(t) \ne \lim_{t \to 14.75^{+}} f(t)$ .

To learn more on piecewise function: brainly.com/question/12561612

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1 year ago

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2 years ago

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Given $a \equiv 1 \pmod{7}$, $b \equiv 2 \pmod{7}$, and $c \equiv 6 \pmod{7}$, what is the remainder when $a^{81} b^{91} c^{27}$

Blizzard [7]

$\begin{cases}a\equiv1\pmod7\\b\equiv2\pmod7\\c\equiv6\pmod7\end{cases}$

$a^{81}\equiv1^{81}\equiv1\pmod7$

$b^{91}\equiv2^{91}\pmod7$

$2^{91}\equiv(2^3)^{30}\times2^1\equiv8^{30}\times2\pmod7$
$8\equiv1\pmod7$
$2\equiv2\pmod7$
$\implies2^{91}\equiv1^{30}\times2\equiv2\pmod7$

$c^{27}\equiv6^{27}\pmod7$

$6^{27}\equiv(-1)^{27}\equiv-1\equiv6\pmod7$

$\implies a^{81}b^{91}c^{27}\equiv1\times2\times6\equiv12\equiv5\pmod7$

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2 years ago