Name 3 different pairs of fractions that have the same product when multiplied

Which is the value of this expression when j = negative 2 and k = negative 1?

Natalija [7]

Answer:

Option A.

Step-by-step explanation:

It is given that

$j=-2, k=-1$

The given expression is

$\left(\dfrac{jk^{-2}}{j^{-1}k^{-3}}\right)^3$

Substitute the given values in the above expression.

$\left(\dfrac{(-2)(-1)^{-2}}{(-2)^{-1}(-1)^{-3}}\right)^3$

$\Rightarrow \left(\dfrac{(-2)(1)}{\dfrac{1}{-2}(-1)}\right)^3$

$\Rightarrow \left(\dfrac{-2}{\dfrac{1}{2}}\right)^3$

$\Rightarrow \left(-4\right)^3$

$\Rightarrow -64$

Therefore, the correct option is A.

3 0

3 years ago

Read 2 more answers

Graph the image of trapezoid ABCD after a translation 3 units left .

Kay [80]

Answer:

A: (-8, -1)

B: (2, -1)

C: (2, 5)

D: (-6, 5)

Step-by-step explanation:

If the trapezoid ABCD was able to translate 3 units to the left, we'll only need to calculate for all the input values!

Trapezoid in present:

A - -5

B - 5

C - 5

D - -3

Now move the units to the left!

Trapezoid in the future:

A - -8

B - 2

C - 2

D - -6

Then add the y values into the units

A: (-8, -1)

B: (2, -1)

C: (2, 5)

D: (-6, 5)

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7 0

3 years ago

Identify the vertex, axis of symmetry, minimum or maximum, domain, and range of the function f(

alekssr [168]

Identify the vertex, axis of symmetry, minimum or maximum, domain, and range of the function ()=−(+)^−

Answer:

vertex = (-4, -5)

Axis of symmetry = -4

use the (-4, -5) to find the minimum value

$Domain = ( - \infty, \infty ) , [ x | x\ is\ real ]\\\\Range = [ -5, \infty ), y\geq -5$

Solution:

Given function is:

$f(x) = (x+4)^2 - 5$

The equation in vertex form is given as:

$y = a(x-h)^2+k$

Where, (h, k) is constant

On comparing give function with vertex form,

h = -4

k = -5

Vertex is (-4 , -5)

Axis of symmetry : x co-ordinate of vertex

Thus, axis of symmetry = -4

The coefficient of x^2 is positive in given function.

Thus the vertex point will be a minimum

$Minimum\ value = f(\frac{-b}{a})$

$f(x) = x^2 + 8x + 16 - 5\\\\f(x) = x^2 + 8x + 11$

$f(x) = ax^2+bx+c$

On comparing,

a = 1

b = 8

$x = \frac{-b}{2a} = \frac{-8}{2 \times 1} = -4$

$f(-4) = (-4)^2 + 8(-4) + 11 = 16 - 32 + 11 = -5$

Thus, use the (-4, -5) to find the minimum value

Domain and range

$f(x) = (x+4)^2 - 5$

The domain is the input values shown on the x-axis

The range is the set of possible output values f(x)

Therefore,

$Domain = ( - \infty, \infty ) , [ x | x\ is\ real ]\\\\Range = [ -5, \infty ), y\geq -5$

4 0

3 years ago

(this is separate)

leva [86]

Answer: 2/1

Step-by-step explanation:

3 0

3 years ago

Convert 232°F to Celsius. 55°C 95°C 100°C 110°C

Ilya [14]

232- 449.6
55- 135
95- 203
100- 212
110- 230

3 0

3 years ago