What is the solution of 3x+y=2 & 4y=12-12x?

I need to find the answer in math for2×2/3 what does it equal

Nataly_w [17]

Here u go the answer is 4/3

6 0

3 years ago

Need help with inscribed angles If M

Bogdan [553]

Given:

In the given circle O, BC is diameter, OA is radius, DC is a chord parallel to chord BA and $m\angle BCD=30^\circ$ .

To find:

The $m\angle AOB$ .

Solution:

If a transversal line intersect two parallel lines, then the alternate interior angles are congruent.

We have, DC is parallel to BA and BC is the transversal line.

$\angle OBA\cong \angle BCD$ [Alternate interior angles]

$m\angle OBA=m\angle BCD$

$m\angle OBA=30^\circ$

In triangle AOB, OA and OB are radii of the circle O. It means OA=OB and triangle AOB is an isosceles triangle.

We know that base angles of an isosceles triangle are congruent.

$\angle OAB\cong \angle OBA$ [Base angles of an isosceles triangle]

$m\angle OAB=m\angle OBA$

$m\angle OAB=30^\circ$

In triangle AOB,

$m\angle OAB+m\angle OBA+m\angle AOB=180^\circ$

$30^\circ+30^\circ+m\angle AOB=180^\circ$

$60^\circ+m\angle AOB=180^\circ$

$m\angle AOB=180^\circ-60^\circ$

$m\angle AOB=120^\circ$

Therefore, the measure of angle AOB is 120 degrees.

6 0

2 years ago

Use integration by parts to find the integrals in Exercise.<br> ∫(4x-12)e-8x dx.

stepladder [879]

Answer:

$e(2x^{2} -12x)-4x^{2}+C$

Step-by-step explanation:

We have been given an indefinite integral as $\int \left(4x-12\right)e-8x\:dx$ . We are asked to find the given integral.

Let us solve our given problem.

$\int \left(4x-12\right)e\:dx-\int 8x\:dx$

Take out constant:

$e\int \left(4x-12\right)\:dx-8\int x\:dx$

$e(\int 4x\:dx -\int 12\right\:dx)-8\int x\:dx$

$e(\frac{4x^{1+1}}{2} -12x)-8*\frac{x^{1+1}}{1+1}+C$

$e(\frac{4x^{2}}{2} -12x)-8*\frac{x^{2}}{2}+C$

$e(2x^{2} -12x)-4x^{2}+C$

Therefore, our required integral would be $e(2x^{2} -12x)-4x^{2}+C$ .

5 0

2 years ago

Find a formula for the described function. A rectangle has perimeter 16 m. Express the area A of the rectangle as a function of

Alex17521 [72]

Part A:

Let the length of one of the sides of the rectangle be L, then the length of the other side is obtained as follow.

Let the length of the other side be x, then

$2(L+x)=16 \\ \\ \Rightarrow L+x=8 \\ \\ \Rightarrow x=8-L$

Thus, if the length of one of the side is x, the length of the other side is 8 - L.

Hence, the area of the rectangle in terms of L is given by

$L(8 - L) = 8L-L^2$

Part B:

To find the domain of A

Recall that the domain of a function is the set of values which can be assumed by the independent variable. In this case, the domain is the set of values that L can take.

Notice that the length of a side of a rectangle cannot be negative or 0, thus L cannot be 8 as 8 - 8 = 0 or any number greater than 8.

Hence the domain of the area are the set of values between 0 and 8 not inclusive.

Therefore,

$dom(Area)=0\ \textless \ L\ \textless \ 8 \ or \ (0, 8)$

3 0

3 years ago

Plz help with this math. Can't find the area of the triangular faces.

Zina [86]

First of all, you have to find the area of both triangles:
$\frac{bh}{2}$
$\frac{4*4}{2}$
Or just 16 because there are 2 of the same triangles.

Now you have to find the area of the 3 rectangles.

The two that are in the front are 4*3 (l*h) or 12*2 (because there are 2 congruent rectangles. The area of those rectangles is 24 square mm.

Now you find the area of the back rectangle:
5.7*3 = 17.1

Finally, you add all the found numbers to figure out the surface area.
17.1 + 16 + 24 = 57.1 square millimeters.

Hope this helped,
Loafly

4 0

3 years ago