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3 years ago

What’s the correct answer for this?

Mathematics

2 answers:

Maurinko [17]3 years ago

7 0

Answer:

Step-by-step explanation:

Shalnov [3]3 years ago

5 0

Answer:

P=19.2 ft

Step-by-step explanation:

ACCORDING TO THE THEOREM,

BC=2(DE)

So BC =2(3.2)

BC=6.4 ft

Now

AB=2(EF)

AB=2(4)

AB=8 ft

Now

AC = 2(DF)

AC= 2(2.4)

AC=4.8 ft

Now we're gonna find the perimeter which is sum of all sides

Perimeter=4.8+8+6.4

P= 19.2 ft

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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

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

To find the value of b,

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 $78.42 \mathrm{in}^{2}$

6 0

4 years ago

Read 2 more answers

Use lagrange multiplier techniques to find the local extreme values of f(x, y) = x2 − y2 − 2 subject to the constraint x2 + y2 =

dexar [7]

Given $f(x,\ y)=x^2-y^2-2$ subject to the constraint $x^2+y^2=16$

Let $g(x,\ y)=x^2+y^2$ .

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$\nabla f(x,\ y)=\langle2x,-2y\rangle$ and $\nabla g(x,\ y)=\langle2x,2y\rangle$

By Lagrange's theorem, there is a number $\lambda$ , such that

$\langle2x,-2y\rangle=\lambda\langle2x,2y\rangle=\langle2\lambda x,2\lambda y\rangle$

$\lambda=\pm1$

It can be seen that $f(x,\ y)=x^2-y^2-2$ has local extreme values at the given region.

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3 years ago

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topjm [15]

Answer:

The graph in the attached figure

Step-by-step explanation:

we have

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