The pair of triangles below have

What is the solution to 15 = d - 1 ?

Fofino [41]

Answer:

d = 16

Step-by-step explanation:

15 = d-1

Add 1 to each side

15 +1 =d-1+1

16 =d

8 0

3 years ago

Mike observed that 75% of the students of a school liked skating. If 35 students of the school did not like skating, the number

igor_vitrenko [27]

100-75=25% did not like
25%=35 students how about 75%
75 x 35 divided by 25=105 students
105

3 0

3 years ago

Read 2 more answers

1 What is the equation of the line parallel to 8x - 16y= -14 and passing through (3, 0)?

andreev551 [17]

Answer:

8x -16y = 24

Step-by-step explanation:

8x - 16y= -14

A parallel line (i.e. same slope) will be: 8x - 16y= c

We have to find the value of "c".

If it passes through (x, y) = (3, 0) ⇒ 24 - 0 = c,

so equation is 8x -16y = 24

7 0

3 years ago

Male and female high school students reported how many hours they worked each week in summer jobs. The data is represented in th

Effectus [21]

Answer:

60 percent males 40 percent females

Step-by-step explanation:

there are more males in the world than female.

4 0

3 years ago

Read 2 more answers

A circle is translated 4 units to the right and then reflected over the x-axis. Complete the statement so that it will always be

irga5000 [103]

Answer:

The statement is now presented as:

$\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}$

In other words, this mathematical statement can be translated as:

There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r.

Step-by-step explanation:

Let $C = (h,k)$ the coordinates of the center of the circle, which must be transformed into $C'=(h', k')$ by operations of translation and reflection. From Analytical Geometry we understand that circles are represented by the following equation:

$(x-h)^{2}+(y-k)^{2} = r^{2}$

Where $r$ is the radius of the circle, which remains unchanged in every operation.

Now we proceed to describe the series of operations:

1) Center of the circle is translated 4 units to the right (+x direction):

$C''(x,y) = C(x, y) + U(x,y)$ (Eq. 1)

Where $U(x,y)$ is the translation vector, dimensionless.

If we know that $C(x, y) = (h,k)$ and $U(x,y) = (4, 0)$ , then:

$C''(x,y) = (h,k)+(4,0)$

$C''(x,y) =(h+4,k)$

2) Reflection over the x-axis:

$C'(x,y) = O(x,y) - [C''(x,y)-O(x,y)]$ (Eq. 2)

Where $O(x,y)$ is the reflection point, dimensionless.

If we know that $O(x,y) = (h+4,0)$ and $C''(x,y) =(h+4,k)$ , the new point is:

$C'(x,y) = (h+4,0)-[(h+4,k)-(h+4,0)]$

$C'(x,y) = (h+4, 0)-(0,k)$

$C'(x,y) = (h+4, -k)$

And thus, $h' = h+4$ and $k' = -k$ . The statement is now presented as:

$\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}$

In other words, this mathematical statement can be translated as:

There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r.

4 0

3 years ago