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4 years ago

8.x + 2y = -12 and 5x - y = 4.

Mathematics

1 answer:

Inessa05 [86]4 years ago

5 0

Answer:

Step-by-step explanation:

Graph both equations

The solutions are the coordinates of the intetsection point between the lines.

The coordinates of the intersection point are approximatively (-0.23,-5.12)

So:

● x = -0.23

● y = -5.12

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Given that AXB is complementary to both CYD and FZE, and mAXB = 36°, what is mCYD + mFZE?

anyanavicka [17]

Complementary angles always equal 90°
So here is how to find what mCYD + mFZE equal
mCYD + mFZE + mAXB
mCYD + mFZE + 36° = 90°
Subtract
mCYD + mFZE = 54°

7 0

3 years ago

Simplify (x^3)(x^-2)

Charra [1.4K]

Hey there!

When multiplying variables with the same degree, be sure to add the two degrees being multiplied.

For example;

x * x; both of these terms have the degree of "1" as they are "x^1". Since we are multiplying them, we need to add their degrees.
x^1 * x^1 = x^2

Now that we've broken down what we are doing, let's apply this new knowledge to your problem.

(x^3)(x^-2); Remember to just add the degrees; -2 + 3 = 1
This means our product is x^1, or just x.

I hope this helps!

8 0

4 years ago

Solve for b. r = (b-5)m

alukav5142 [94]

Answer:

b=5m+r/m

Step-by-step explanation:

Let's solve for b.

r=(b−5)(m)

Step 1: Flip the equation.

bm−5m=r

Step 2: Add 5m to both sides.

bm−5m+5m=r+5m

bm=5m+r

Step 3: Divide both sides by m.

bm/m=5m+r/m

b=5m+r/m

Answer:

b=5m+r/m

6 0

3 years ago

Read 2 more answers

Please help, i have to finish this right now

Naddik [55]

1st one: linear
2nd one: linear
3rd one: nonlinear

3 0

3 years ago

Read 2 more answers

How to solve this question

VMariaS [17]

The variance method is as follows.

-Sum the squares of the values in data set, and then divide by the number of values in data set
- From that, subtract the square of the mean (add all values and divide by number of values in the data set)

Our variance is

 $\displaystyle\sigma^2 = \frac{2^2 + 5^2 + m^2}{3} - \left(\frac{2 + 5 + m}{3}\right)^2$

Since variance has to be 14, we set $\sigma^2 = 14$ and solve for m

$14= \frac{4 + 25 + m^2}{3} - \left(\frac{7 + m}{3}\right)^2\ \Rightarrow \\ \\
14 = \frac{29}{3} + \frac{1}{3}m^2 - \frac{1}{9}(7+m)^2 \\ \\
14 = \frac{29}{3}+ \frac{1}{3}m^2 - \frac{1}{9}(49 + 14m + m^2) \\ \\
14 = \frac{29}{3}+ \frac{1}{3}m^2 - \frac{49}{9}- \frac{14}{9}m- \frac{1}{9}m^2 \\ \\
0 = \frac{-88}{9} -\frac{14}{9}m + \frac{2}{9}m^2
$

quadratic formula

$m = \displaystyle\frac{-b \pm \sqrt{b^2 -4ac}}{2a} \\
m = \frac{-(-\frac{14}{9}) \pm \sqrt{\left(-\frac{14}{9}\right)^2 - 4(2/9)(-88/9)} }{2(2/9)} \\
m = \frac{\frac{14}{9} \pm \sqrt{ \frac{196}{81} + \frac{704}{81} } }{\frac{4}{9} } \\
m = \frac{\frac{14}{9} \pm \sqrt{ \frac{900}{81} } }{\frac{4}{9} } \\
m = \frac{\frac{14}{9} \pm \sqrt{ \frac{100}{9} } }{\frac{4}{9} } \\
m = \frac{\frac{14}{9} \pm \frac{10}{3} }{\frac{4}{9} } \\
m = 11, -4$

-4 doesnt' work as it is not a positive integer

m = 11

5 0

3 years ago