can someone please help me with this? I know it’s simple but I’m having a really hard time doing it .

Help answer these two pretty please

4vir4ik [10]

Answer:

none of these triangles are similar

Step-by-step explanation:

for a triangle to be similar, they have to have the same angle measures

4 0

3 years ago

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Is 3/5 equivalent to 6/8

WARRIOR [948]

No because 3*2= 6 while 5*2=10 so if it was equivalent it would of been 3/5 to 6/10

4 0

3 years ago

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Given the functions f(x) = − 4x − 1 and g(x) = 2x + 4, which operation results in the smallest coefficient on the x term?

Sedbober [7]

F(x)=-4x-1 if you sub 3 into it you get 3(-4)-1 which is -13 and 2x+4=10

4 0

3 years ago

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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

COSO ---

Vsevolod [243]

Answer:

• zero: -4, -4/3, 7

• positive: -4 < x < -4/3 . . . or 7 < x

• negative: x < -4 . . . or -4/3 < x < 7

Step-by-step explanation:

Zeros of the function are at x=-4, -4/3, +7. These are the values that make each of the individual factors be zero. For example, x-7=0 when x=7.

The function will be negative for x-values left of an odd number of zeros. It will be positive for x-values left of an even number of zeros (including left of no zeros, which is to say right of all zeros). This is because the sign of the factor giving rise to the zero changes for x-values on either side of that zero. (This is not true for zeros with even multiplicity, as the sign does not change at those.)

3 0

3 years ago