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3 years ago

Is a triangle with the angles 80, 40, and 40 an unique triangle

Mathematics

2 answers:

Anni [7]3 years ago

8 0

Answer:

yes

Step-by-step explanation:

KatRina [158]3 years ago

5 0

Answer:

yes

Step-by-step explanation:

hope this helps

You might be interested in

Solve for m.<br> m/9 + 2/3 =7/3

Nuetrik [128]

Answer:

m= 15

Step-by-step explanation:

$\frac{m}{9} + \frac{2}{3} = \frac{7}{3}$

To solve for m, start by moving the constants (numbers that are not attached to any variable) to the other side of the equation.

$\frac{m}{9} = \frac{7}{3} - \frac{2}{3}$

Simplify:

$\frac{m}{9} = \frac{5}{3}$

Multiply both sides by 9:

$m = \frac{5}{3} \times 9$

Crossing out 3 from the denominator and from 9:

m= 5(3)

m= 15

8 0

2 years ago

Plz help ASAP!! Explain your answer! I will mark at brainliest!!! And don’t copy anybody else’s answer

Klio2033 [76]

Answer:

No, it is not a square

Step-by-step explanation:

If one wall is 19", that would mean the wall perpendicular to this wall is also 19" (in fact all of the walls would be 19"!) If this was a square, then the diagonal we draw at 20.62" would serve as the hypotenuse of a right triangle. One wall would serve as a leg, and another wall as another leg. If this is a square, then the Pythagorean's Theorem would be satisfied when we plug in the 2 wall measures for a and b, and the diagonal for c:

$19^2+19^2=20.62^2$

We need to see if this is a true statement. If the left side equals the right side, then the 2 legs of the right triangle are the same length, and the room, then is a square.

361 + 361 = 425.1844

Is this true? Does 722 = 425.1844? Definitely not. That means that the room is not a square.

8 0

3 years ago

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$