What is the range? 5 7 10 7 3 10 9 1 1 6

2 answers:

8 0

The range is 9.

You can find range by subtracting the lowest number from the highest number.

10-1=9

Hope this helps ✨❤️

fenix001 [56]3 years ago

4 0

Answer:

Step-by-step explanation:

10-1=9 thats how you find the range you do the highest number then the lowest number after that you subtract it

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What is the angle of BTC, Angle TBC,Angle TCB

Goryan [66]

By geometric and algebraic properties the angles BTC, TBC and TBC from the triangle BTC are 128°, 26° and 26°, respectively.

<h3>How to determine the angles of a triangle inscribed in a circle</h3>

According to the figure, the triangle BTC is inscribed in the circle by two points (B, C). In this question we must make use of concepts of diameter and triangles to determine all missing angles.

Since AT and BT represent the radii of the circle, then the triangle ABT is an isosceles triangle. By geometry we know that the sum of internal angles of a triangle equals 180°. Hence, the measure of the angles A and B is 64°.

The angles ATB and BTC are supplmentary and therefore the measure of the latter is 128°. The triangle BTC is also an isosceles triangle and the measure of angles TBC and TCB is 26°.

By geometric and algebraic properties the angles BTC, TBC and TBC from the triangle BTC are 128°, 26° and 26°, respectively.

To learn more on triangles, we kindly invite to check this verified question: brainly.com/question/2773823

8 0

2 years ago

Please help if you can actually do this and not for the points

Tcecarenko [31]

The answer you are looking for is

7 0

3 years ago

Need help simplifying this

diamong [38]

The simplified answer is $\frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}$ .

Step-by-step explanation:

$$\frac{3 y+2 x}{z+2 x}-\frac{2 y-3 x}{3 x+y}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}$

To add or subtract denominators of the fraction must be same.

If it is not the same, we must take LCM of the denominators. and so we can add the fractions.

To make the denominator same multiply the 1st term $(\frac{3x+y}{3x+y})$ and 2nd term by $(\frac{z+2x}{z+2x})$

= $\frac{(3 y+2 x)(3 x+y)}{(z+2 x)(3 x+y)}-\frac{(2 y-3 x)(z+2 x)}{(3 x+y)(z+2 x)}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}$

LCM of the denominators is 6x²+ 3xz + 2xy +yz.

Multiply the factors in the numerator.

= $\frac{\left(6 x^{2}+3 y^{2}+11 x y\right)}{(z+2 x)(3 x+y)}-\frac{\left(2 y z+4 x y-3 x z-6 x^{2}\right)}{(3 x+y)(z+2 x)}-\frac{2 z y+6 x z}{6 x^{2}+3 x z+2 x y+y z}$