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Nitella [24]
4 years ago
11

Is the equation true or false when x = 4? (3−x28)⋅12=3x(3−x28)⋅12=3x true false

Mathematics
1 answer:
KatRina [158]4 years ago
6 0

Answer:

It’s true

Step-by-step explanation:

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2. (08.05 HC)
4vir4ik [10]

Answer:

option 3

Step-by-step explanation:

not sure if right

8 0
3 years ago
Graph a line that contains the point ( 3 , -6) and has a slope of 1/2
irina [24]
You need to physically draw that on a graph
3 0
3 years ago
The domain is ?<br> The range is ?
DIA [1.3K]

This is a quadratic equation, which is also a polynomial.  Polynomials all have the domain (-infinity, +infinity).

As for the range:  You can see from the graph that the smallest y-value is -2.  Thus, the range is [-2, infinity).

4 0
3 years ago
Last week a candle store received $515.90 for selling 30 candles. Small candles sell for $11.25 and large candles sell for $22.4
alex41 [277]

Answer:

16

Step-by-step explanation:

x+y=30  solve for x   x=30-y

substitute x with 30-y and solve

11.25x+22.4y=515.9

11.25(30-y)+22.4y=515.9

337.5+11.15y=515.9

11.15y=178.4

y=16

since we're solving for just large candles it's 16

if need small cangle number 30-16=14

8 0
3 years ago
Please help me solve this answer it all for me please
mestny [16]

Answer:

The slopes are

m1=\dfrac{2}{5}, m2=-\dfrac{5}{2}

Therefore, the equations are equations of <u>  Perpendicular Lines .</u>

Step-by-step explanation:

Given:

y=\dfrac{2}{5}\times x + 1    ......................Equation ( 1 )

5x+2y=-4\\\\\therefore y = \dfrac{-5}{2}\times x-2   ..............Equation ( 2 )

To Find:

Slope of equation 1 = ?

Slope of equation 2 = ?

Solution:

On comparing with slope point form

y=mx+c

Where,

m = Slope

c = y-intercept

We get

Step 1.

Slope of equation 1 = m1 = \dfrac{2}{5}

Step 2.

Slope of equation 1 = m2 = -\dfrac{5}{2}

Step 3.

Product of Slopes = m1 × m2 = \dfrac{2}{5}\times -\dfrac{5}{2}=-1

Product of Slopes = m1 × m2 = -1

Which is the condition for Perpendicular Lines

The slopes are

m1=\dfrac{2}{5},m2=-\dfrac{5}{2}

Therefore, the equations are equations of <u>  Perpendicular Lines . </u>

4 0
3 years ago
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