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3 years ago

11

The graph of h (x) = StartAbsoluteValue x minus 10 EndAbsoluteValue + 6 is shown. On which interval is this graph increasing? (–

∞, 6) (–∞, 10) (6, ∞)

2 answers:

Andrew [12]3 years ago

7 0

Answer:

D. (10, Infinity)

Step-by-step explanation:

Got it right on edge.

Andrew [12]3 years ago

3 0

Answer:

(10, ∞)

Step-by-step explanation:

i got you

big brain

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4 years ago

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3 years ago

In parallelogram ABCD, AD = 12 in, m∠C = 46º, m∠DBA = 72º. Find the area of ABCD

dmitriy555 [2]

Answer:

The area of parallelogram ABCD is $78.42 \mathrm{in}^{2}$

Explanation:

Given:

AD = 12 in

$m \angle C=46^{\circ}$

$m \angle D B A=72^{\circ}$

To Find:

The area of parallelogram ABCD=?

Solution:

When we construct the parallelogram with the given data, we get a parallelogram formed by 12 cm as one side and an angle with 46 degrees.

The area of the parallelogram can be calculated by $a b * \sin (a n g l e)$

Substituting the value of a=12 we have

$\text { Area of parallelogram }=12 * \text { bsin } 46$

<u>To find the value of b, </u>

We know that area of a triangle can be expressed as,

$\text { Area of triangle }=(A b / 2) \sin (\text {angle})$

So,

$(12 * B D / 2) * \sin 46=(A B * B D / 2) * \sin 72$

Cancelling BD and 2 on both sides we get,

$12 * \sin 46=A B * \sin 72$

$A B=12 * \frac{\sin 46}{\sin 72}$

Therefore,

$b=\frac{12 \sin 46}{\sin 72}$

Substituting the value of b,

$=12 *\left(\frac{12 \sin 46}{\sin 72}\right) * \sin 46$

=78.42

So the area of the parallelogram is $78.42 \mathrm{in}^{2}$

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