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2 years ago

What is the radius of the circle described by the equation (x-6)* +(y - 4)- 64 = 0?

Mathematics

1 answer:

daser333 [38]2 years ago

7 0

Answer:

d) 8

Step-by-step explanation:

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Given congruent triangles name the corresponding sides and corresponding angles

Alborosie

Answer:

See explanation

Step-by-step explanation:

Given $\triangle ABC\cong \triangle ADC$

According to the order of the vertices,

side AB in triangle ABC (the first and the second vertices) is congruent to side AD in triangle ADC (the first and the second vertices);

side BC in triangle ABC (the second and the third vertices) is congruent to side DC in triangle ADC (the second and the third vertices);

side AC in triangle ABC (the first and the third vertices) is congruent to side AC in triangle ADC (the first and the third vertices);

angle BAC in triangle ABC is congruent to angle DAC in triangle ADC (the first vertex in each triangle is in the middle when naming the angles);

angle ABC in triangle ABC is congruent to angle ADC in triangle ADC (the second vertex in each triangle is in the middle when naming the angles);

angle BCA in triangle ABC is congruent to angle DCA in triangle ADC (the third vertex in each triangle is in the middle when naming the angles);

6 0

3 years ago

The triangle and square shown have the same perimeter. what expression represents of the triangle in terms of x? Give your awnse

Rashid [163]

Answer:

brainly.com/question/17506481

Step-by-step explanation:

5 0

3 years ago

1.) Simplify: y = -8 + 6(x-1)

swat32

8+6=14. (X-1=1x)

14•1x

6 0

2 years ago

Find the roots of the equation<br> x ^ 2 + 3x-8 ^ -14 = 0 with three precision digits

scoray [572]

Answer:

Step-by-step explanation:

Given quadratic equation:

$x^{2} + 3x - 8^{- 14} = 0$

The solution of the given quadratic eqn is given by using Sri Dharacharya formula:

$x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}$

The above solution is for the quadratic equation of the form:

$ax^{2} + bx + c = 0$

$x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}$

From the given eqn

a = 1

b = 3

c = $- 8^{- 14}$

Now, using the above values in the formula mentioned above:

$x_{1, 1'} = \frac{- 3 \pm \sqrt{3^{2} - 4(1)(- 8^{- 14})}}{2(1)}$

$x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})})$

$x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})} - 3)$

Now, Rationalizing the above eqn:

$x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(- 8^{- 14})} - 3)\times (\frac{\sqrt{9 - 4(- 8^{- 14})} + 3}{\sqrt{9 - 4(- 8^{- 14})} + 3}$

$x_{1, 1'} = \frac{1}{2}.\frac{(\pm {9 - 4(- 8^{- 14})^{2}} - 3^{2})}{\sqrt{9 - 4(- 8^{- 14})} + 3}$

Solving the above eqn:

$x_{1, 1'} = \frac{2\times 8^{- 14}}{\sqrt{9 + 4\times 8^{-14}} + 3}$

Solving with the help of caculator:

$x_{1, 1'} = \frac{2\times 2.27\times 10^{- 14}}{\sqrt{9 + 42.27\times 10^{- 14}} + 3}$

The precise value upto three decimal places comes out to be:

$x_{1, 1'} = 0.758\times 10^{- 14}$

5 0

3 years ago

"For each power, write an equivalent multiplication expression with two factors"

Eva8 [605]

3 0

2 years ago