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kherson [118]
3 years ago
14

What has no congruent sides and no congruent angles?

Mathematics
2 answers:
____ [38]3 years ago
6 0
A circle

Hope I helped!

~ Zoe
Mademuasel [1]3 years ago
3 0
Hey there!

The correct answer to your question is a circle.

A circle doesn't have any congruent sides or congruent angles, meaning they don't have sides or angles that point in the same direction.

Hope this helps you.
Have a great day!
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HELP{
sertanlavr [38]

Answer: The correct conclusion is(B) The functions f(x) and g(x) are reflections over the y-axis.

Step-by-step explanation: Two functions f(x) and g(x) are given as follows:

f(x)=2^x,~~~~~~~g(x)=\left(\dfrac{1}{2}\right)^x.

We know that if f(-x) = g(x), then the functions are reflections over Y-axis and if - f(x) = g(x), then the functions are reflections over X-axis.

We have,

f(-x)=2^{-x}=\left(\dfrac{1}{2}\right)^x=g(x),\\\\-f(x)=-2^x\neq g(x).

So, the function g(x) is a reflection of f(x) over Y-axis.

The graph of f(x) and g(x) are drawn in the attached file. From there, it is clear that the functions are reflections over Y-axis, not reflections over X-axis.

So, options (A) is incorrect and option (B) is correct.

From the table, we have

f(-2)=\dfrac{1}{4},~~f(-1)=\dfrac{1}{2},~~f(0)=1,~~f(1)=2,~~f(2)=4,\\\\g(-2)=4,~~g(-1)=2,~~g(0)=1,~~g(1)=\dfrac{1}{2},~~g(2)=\dfrac{1}{4}.

So, as the value of 'x' increases, the value of f(x) increases and value of y(x) decreases.

Therefore, f(x) is an increasing function and g(x) is a decreasing function. So, option (C) is incorrect.

Also, we have

f(0)=g(0)=1.

So, both the functions have same initial value. So, option (D) is also incorrect.

Thus, the correct conclusion is (B) The functions f(x) and g(x) are reflections over the y-axis.

4 0
3 years ago
Read 2 more answers
Point GG is located at (-4,2)(−4,2) on the coordinate plane. Point GG is reflected over the xx-axis to create point G'G ′ . What
Taya2010 [7]

Answer: (4,2)

Step-by-step explanation:

When a point is reflected over the x axis, only the x value changes to the opposite sign (positive or negative). The y value doesn’t change.

7 0
3 years ago
Un padre tiene actualmente cuatro veces la edad de su hijo. Cuando pasen 5 años, su edad será solo tres veces superior. ¿Qué eda
julsineya [31]

Answer:

I. Hijo, h = 10 años

II. Padre, P = 40 años

Step-by-step explanation:

  • Sea la edad del padre P.
  • Sea la edad del hijo h.

Traduciendo el problema verbal a una expresión algebraica, tenemos;

P = 4h .....ecuación 1.

P + 5 = 3(h + 5) ........ecuación 2.

Simplificando aún más, tenemos;

P + 5 = 3h + 15

P = 3h + 15 - 5

F = 3h + 10 ......ecuación 3.

Sustituyendo la ecuación 1 en la ecuación 3;

4h = 3h + 10

4h - 3h = 10

<em>h = 10 años </em>

A continuación, encontraríamos la edad del padre;

P = 4h

P = 4 * 10

<em>P = 40 años</em>

4 0
3 years ago
Point A is located at (5, 10) and point B is located at (20, 25).
Alla [95]
Let's say the point is C, so C partitions AB into two pieces, where AC is at a ratio of 3 and CB is at  a ratio of 7, thus 3:7,

\bf \left. \qquad  \right.\textit{internal division of a line segment}&#10;\\\\\\&#10;A(5,10)\qquad B(20,25)\qquad&#10;\qquad 3:7&#10;\\\\\\&#10;\cfrac{AC}{CB} = \cfrac{3}{7}\implies \cfrac{A}{B} = \cfrac{3}{7}\implies 7A=3B\implies 7(5,10)=3(20,25)\\\\&#10;-------------------------------\\\\&#10;{ C=\left(\cfrac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \cfrac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)}\\\&#10;-------------------------------

\bf C=\left(\cfrac{(7\cdot 5)+(3\cdot 20)}{3+7}\quad ,\quad \cfrac{(7\cdot 10)+(3\cdot 25)}{3+7}\right)&#10;\\\\\\&#10;C=\left( \cfrac{35+60}{10}~~,~~\cfrac{70+75}{10} \right)\implies C=\left(\cfrac{95}{10}~~,~~\cfrac{145}{10}  \right)&#10;\\\\\\&#10;C=\left( \cfrac{19}{2}~~,~~\cfrac{29}{2} \right)\implies C=\left( 9\frac{1}{2}~~,~~14\frac{1}{2}  \right)
8 0
3 years ago
Solve these equations.<br> |x-7|=x-7
Cloud [144]

Answer:

x ≥ 7

Step-by-step explanation:

|x - 7| = x - 7

A. For each absolute, find the intervals

x - 7 ≥ 0     x - 7 < 0

     x ≥ 7           x < 7

If x ≥ 7, |x - 7| = x - 7 > 0.

If x < 7, |x - 7| = x - 7 < 0. No solution.

B. Solve for x < 7

Rewrite |x - 7| = x - 7 as

+(x - 7) = x - 7

      x - 7 = x - 7

-x + 14 = x

        14 = 2x

          x = 7

7 ≮7. No solution

C. Solve for x ≥ 7

Rewrite |x - 7| = x - 7 as

+(x - 7) = x - 7

      x - 7 = x - 7

True for all x.

D. Merge overlapping intervals

No solution or x ≥ 7

⇒ x ≥ 7

The diagram below shows that the graphs of y = |x - 7| (blue) and of y = x - 7 (dashed red) coincide only when x ≥ 7.

6 0
3 years ago
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