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vovikov84 [41]
3 years ago
6

Which equation can compare with 3X plus 4Y equals eight to create a consistent and independent system

Mathematics
1 answer:
Scilla [17]3 years ago
3 0

Answer:

A consistent and independent system is any equation that only has one solution. So an another equation that can compare with this to create one solution is an equation that is perpendicular. The slope will be opposite and reciprocal.

This is the first equation in standard form: 3x + 4y = 8

Point-slope form is: y = -3/4x + 2

The second equation in standard form: y = 4/3x + 2

Point-slope form is: -4x + 3y = 6

As you can see, the slopes are opposite and reciprocal, which creates two perpendicular lines when graphed. You can graph these two lines, and see that they have one solution.

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Work out angle bxc. give a reason for each angle you work out
Rus_ich [418]

Answer:

∠ BXC = 70°

Step-by-step explanation:

∠ XBC and ∠ AXY are corresponding angles and are congruent, then

∠ XBC = 55°

Since XB = XC , then Δ XBC is isosceles and the 2 base angles are congruent.

∠ BXC = 180° - (55 + 55)° ← angle sum of triangle

∠ BXC = 180° - 110° = 70°

7 0
2 years ago
Evaluate this: -10x^0
FinnZ [79.3K]

Answer:

-10.

Step-by-step explanation:

x^0 = 1 so its -10 * 1

= -10.

4 0
3 years ago
Read 2 more answers
I need help. Please
zubka84 [21]

Answer:

Answer is A.

Step-by-step explanation:

If you take the current cords of 'C' and add the cords it gives you: (4.5,-6.5). You end out with (3.5,-9,2). Hope this helps!

5 0
3 years ago
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D<br> Evaluate<br> arcsin<br> (6)]<br> at x = 4.<br> dx
sineoko [7]

Answer:

\frac{1}{2\sqrt{5} }

Step-by-step explanation:

Let, \text{sin}^{-1}(\frac{x}{6}) = y

sin(y) = \frac{x}{6}

\frac{d}{dx}\text{sin(y)}=\frac{d}{dx}(\frac{x}{6})

\frac{d}{dx}\text{sin(y)}=\frac{1}{6}

\frac{d}{dx}\text{sin(y)}=\text{cos}(y)\frac{dy}{dx} ---------(1)

\frac{1}{6}=\text{cos}(y)\frac{dy}{dx}

\frac{dy}{dx}=\frac{1}{6\text{cos(y)}}

cos(y) = \sqrt{1-\text{sin}^{2}(y) }

          = \sqrt{1-(\frac{x}{6})^2}

          = \sqrt{1-(\frac{x^2}{36})}

Therefore, from equation (1),

\frac{dy}{dx}=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}

Or \frac{d}{dx}[\text{sin}^{-1}(\frac{x}{6})]=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}

At x = 4,

\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}

\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}

                   =\frac{1}{6\sqrt{\frac{36-16}{36}}}

                   =\frac{1}{6\sqrt{\frac{20}{36} }}

                   =\frac{1}{\sqrt{20}}

                   =\frac{1}{2\sqrt{5}}

4 0
2 years ago
24-x/x = 1/2 find x please i need help
Aliun [14]

Answer:

No Solution

Step-by-step explanation:

There are no values of  x  that make the equation true.

3 0
2 years ago
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