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Rudiy27
3 years ago
10

Factor (z-5) (z+4)=52​

Mathematics
1 answer:
stich3 [128]3 years ago
4 0

Answer:

\large\boxed{(z-9)(z+8)=0\to z=9,\ z=-8}

Step-by-step explanation:

(z-5)(z+4)=52\qquad\text{use FOIL}:\ (a+b)(c+d)=ac+ad+bc+bd\\\\(z)(z)+(z)(4)+(-5)(z)+(-5)(4)=52\\\\z^2+4z-5z-20=52\qquad\text{subtract 52 from both sides}\\\\z^2-z-72=0\\\\z^2-9z+8z-72=0\\\\z(z-9)+8(z-9)=0\\\\(z-9)(z+8)=0\iff z-9=0\ \vee\ z+8=0\\\\z=9\ \vee\ z=-8

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Sedbober [7]
Sum up the total number of votes.
So, 5+7+8=20 students
3 0
3 years ago
The expression (2x2)3 is equivalent to _____.<br><br>6x2<br><br>6x5<br><br>8x6
mamaluj [8]
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6 0
3 years ago
HELP DUE LIKE RIGHT NOW​
Nimfa-mama [501]

Answer:

The correct answer is x = 17.

Step-by-step explanation:

If EF bisects DEG, this means that angles DEF and angles FEG are congruent, and they each make up half of angle DEG.

Therefore, we can set up the equation:

DEF + FEG = DEG

However, since we know that DEF and FEG represent the same value, we can change this equation into the following:

2(DEF) = DEG

Now, we can substitute in the expressions that we are given:

2(3x+1) = 5x + 19

To simplify, we should first use the distributive property on the left side of the equation.

6x + 2 = 5x + 19

Our next step is to subtract 5x from both sides of the equation.

x + 2 = 19

Finally, we can subtract 2 from both sides of the equation to get x by itself on the left side.

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Hope this helps!

7 0
4 years ago
The endpoints of CD are C(−4, 7) and D(0,−3).
tino4ka555 [31]

Midpoint =(\frac{x_{2}+x_{1}}{2},\frac{y_{2}+y_{1}}{2})

Answer:

Midpoints of line CD are (-2,2).

Step-by-step explanation:

The midpoints is given by

Midpoint =(\frac{x_{2}+x_{1}}{2},\frac{y_{2}+y_{1}}{2})

As given

The endpoints of CD are C(−4, 7) and D(0,−3).

Putting values in the above

Midpoint =(\frac{0-4}{2},\frac{-3+7}{2})

Midpoint =(\frac{-4}{2},\frac{4}{2})

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Therefore midpoints of line CD are (-2,2).

4 0
3 years ago
Read 2 more answers
Please help me...........​
Dominik [7]
I think it’s the 1st one
8 0
2 years ago
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