2x²+x-6=0
(2x-3)(x+2)=0
x=3/2 or -2
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If you would like to eliminate the z-term, you can do this using the following steps:
x + z = 6 /*(-2)
3x + 2z = 1
_____________
-2x -2z = -12
<span>3x + 2z = 1
</span><span>_____________
</span>-2x + 3x = -12 + 1
x = -11
The correct result would be <span>x + z = 6 /*(-2).</span>
H(k) = k^2 - k so we just substitute in 10 for K.
h(10) = 10^2 - 10. 100 - 10 = 90
h(10) = 90.
Answer:
Step-by-step explanation:
∠ADC = 90°
∠ADB + ∠BDC = 90
20 + ∠BDC = 90
∠BDC = 90 - 20 = 70
∠BDC = 70°
16) ∠PSR = ∠PSQ +∠QSR
= 60 + 10
= 70
18) ∠PSR = 130
∠PSQ + ∠QSR = 130
90+ ∠QSR = 130
∠QSR = 130 - 90
∠QSR = 40
19)∠ADC = ∠ADB + ∠BDC
= 120 +20
= 140
20) ∠PSR = 125
Log4(x(x-3))=Log4(-7x+21)
Log4(X the power of two -3x)=Log4(-7x21)
X the power of two -3x=-7x+21
x=3,x=-7.