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3 years ago

Which of the following are valid names for the given triangle? Check all that apply.

Mathematics

2 answers:

Georgia [21]3 years ago

7 0

Answer:

A, D and E are correct options.

Step-by-step explanation:

We are given,

The vertices of the triangle are A, X and T.

While writing the names of a triangle, only the vertices are used.

So, we have that the possible names of the given triangle are,

ΔTAX ΔATX ΔXAT

ΔTXA ΔAXT ΔXTA

Then, according to the options, the valid names for the triangle are,

A. ΔTAX

D. ΔXTA

E. ΔAXT

vfiekz [6]3 years ago

6 0

When naming triangles we name them by the letters on each corner so only answers with T, A, and X in them are correct, so TAX, XTA, and AXT

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15 is 10 percent of what number

aksik [14]

Your answer is 1.5 if you multiply 15 by 10% you get 1.5 and if you ivide 1.5 by 10% you get 15 so you can find your answer and check it
hope this helps have a nice day.

4 0

3 years ago

Can someone plz help me with #2, #3, #4, #14

Drupady [299]

#2 is 5
#3 is 18.9
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8 0

3 years ago

HELP ON THIS PROBLEM!! Solve the following 0deg< x< 360deg: cos x= cos 2x

blsea [12.9K]

Answer:

$x=0^\circ,120^\circ,240^\circ,360^\circ$

Step-by-step explanation:

Use identities to set the equation up as a quadratic

$\cos x=\cos 2x;\:0^\circ \leq x \leq 360^\circ\\\\\cos x=2\cos^2 x-1\\\\0=2\cos^2 x-\cos x-1$

Make the substitution u=cos(x) and solve the quadratic

$0=2u^2-u-1\\\\0=(2u+1)(u-1)$

$\displaystyle 0=2u+1\\\\0=2\cos x+1\\\\-1=2\cos x\\\\-\frac{1}{2}=\cos x\\\\120^\circ,240^\circ=x$

$0=u-1\\\\0=\cos x-1\\\\1=\cos x\\\\0^\circ,360^\circ=x$

Hence, $x=0^\circ,120^\circ,240^\circ,360^\circ$

7 0

2 years ago

How to solve 14x + 2 = 6 (x - 0)

Sergeu [11.5K]

Answer:

Has not solution.

Step-by-step explanation:

14x÷2 = 6(x - 0)

14x÷2 = 6*x + 6*-0

14x÷2 = 6x - 0

14x = 2*6x

14x ≠ 12x

then:

This equation has not solution.

8 0

3 years ago

Solving separable differential equation DY over DX equals xy+3x-y-3/xy-2x+4y-8

Ivanshal [37]

It looks like the differential equation is

$\dfrac{dy}{dx} = \dfrac{xy + 3x - y - 3}{xy - 2x + 4y - 8}$

Factorize the right side by grouping.

$xy + 3x - y - 3 = x (y + 3) - (y + 3) = (x - 1) (y + 3)$

$xy - 2x + 4y - 8 = x (y - 2) + 4 (y - 2) = (x + 4) (y - 2)$

Now we can separate variables as

$\dfrac{dy}{dx} = \dfrac{(x-1)(y+3)}{(x+4)(y-2)} \implies \dfrac{y-2}{y+3} \, dy = \dfrac{x-1}{x+4} \, dx$

Integrate both sides.

$\displaystyle \int \frac{y-2}{y+3} \, dy = \int \frac{x-1}{x+4} \, dx$

$\displaystyle \int \left(1 - \frac5{y+3}\right) \, dy = \int \left(1 - \frac5{x + 4}\right) \, dx$

$\implies \boxed{y - 5 \ln|y + 3| = x - 5 \ln|x + 4| + C}$

You could go on to solve for $y$ explicitly as a function of $x$ , but that involves a special function called the "product logarithm" or "Lambert W" function, which is probably beyond your scope.

8 0

2 years ago