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3 years ago

Knock knock? Whos There? My Points

Mathematics

2 answers:

Luden [163]3 years ago

6 0

Answer:

that was a good one. tanks :)

Blizzard [7]3 years ago

3 0

Answer:

thanks

Step-by-step explanation:

tysmmm ^-^ <<33333

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Geometry help asap please

NemiM [27]

Answer:

99 degrees

Step-by-step explanation:

81+a=180, so

180-81=99

a=99

5 0

3 years ago

Directions attatched

irina1246 [14]

Answer:

x=7
∠GBD=112
∠EFC=112

Step-by-step explanation:

Here, we need to know about the concept of the alternate interior angle theorem. This says that the alternate interior angle of two parallel lines formed with one transversal is equal. For a more graphical explanation, please refer to the attachment below.

--------------------------------------------------------------------------

Given

∠GBD=4x+84

∠EFC=16x

m∠GBD=m∠EFC

Solve

4x+84=16x ⇔ Given

4x+84-4x=16x-4x ⇔ Subtract 4x on both sides

84=12x ⇔ Simplify

84/12=12x/12 ⇔ Divide 12 on both sides

x=7 ⇔ Simplify

Substitute the x value into each angle expression

∠GBD=4x+84=4(7)+84=28+84=112

∠EFC=16x=16(7)=112

Hope this helps!! :)

Please let me know if you have any questions

6 0

3 years ago

What is the slope of the line through (-5,-10) and (-1,5)

Brilliant_brown [7]

The slope is 15/4
(up 15, over 4)
or up 3.75, over 1 if you reduce

6 0

4 years ago

Find the volume v of the described solid s. the base of s is an elliptical region with boundary curve 4x2 + 9y2 = 36. cross-sect

Tasya [4]

$4x^2+9y^2=36\iff\dfrac{x^2}9+\dfrac{y^2}4=1$

defines an ellipse centered at $(0,0)$ with semi-major axis length 3 and semi-minor axis length 2. The semi-major axis lies on the $x$ -axis. So if cross sections are taken perpendicular to the $x$ -axis, any such triangular section will have a base that is determined by the vertical distance between the lower and upper halves of the ellipse. That is, any cross section taken at $x=x_0$ will have a base of length

$\dfrac{x^2}9+\dfrac{y^2}4=1\implies y=\pm\dfrac23\sqrt{9-x^2}$
$\implies \text{base}=\dfrac23\sqrt{9-{x_0}^2}-\left(-\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac43\sqrt{9-{x_0}^2}$

I've attached a graphic of what a sample section would look like.

Any such isosceles triangle will have a hypotenuse that occurs in a $\sqrt2:1$ ratio with either of the remaining legs. So if the hypotenuse is $\dfrac43\sqrt{9-{x_0}^2}$ , then either leg will have length $\dfrac4{3\sqrt2}\sqrt{9-{x_0}^2}$ .

Now the legs form a similar triangle with the height of the triangle, where the legs of the larger triangle section are the hypotenuses and the height is one of the legs. This means the height of the triangular section is $\dfrac4{3(\sqrt2)^2}\sqrt{9-{x_0}^2}=\dfrac23\sqrt{9-{x_0}^2}$ .

Finally, $x_0$ can be chosen from any value in $-3\le x_0\le3$ . We're now ready to set up the integral to find the volume of the solid. The volume is the sum of the infinitely many triangular sections' areas, which are

$\dfrac12\left(\dfrac43\sqrt{9-{x_0}^2}\right)\left(\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac49(9-{x_0}^2)$

and so the volume would be

$\displaystyle\int_{x=-3}^{x=3}\frac49(9-x^2)\,\mathrm dx$
$=\left(4x-\dfrac4{27}x^3\right)\bigg|_{x=-3}^{x=3}$
$=16$

6 0

3 years ago

an average of 4 cups of tea are poured per minute. The lunch will last 2 hours. How many gallons of tea should the caterer bring

Arada [10]

I believe it would be 30 gallons, because there are 4 every minute and 120 minutes in two hours. And 4 cups of 4 in in 16

8 0

3 years ago

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