21 uždavinys23 uždavinys
Sprendimas:
- 0.1* x^2+22.5 =
0
- 0.1* x^2+22.5 = 0$$-0.1\cdot x^{2}+22.5$$ = $$0$$
22.5 = 0+ 0.1* x^2$$22.5$$ = $$0+0.1\cdot x^{2}$$
22.5 = 0.1* x^2$$22.5$$ = $$0.1\cdot x^{2}$$
= x^2$$\frac{22.5}{0.1}$$ = $$x^{2}$$ 225 = x^2$$225$$ = $$x^{2}$$
saknis(225) = x$$\sqrt {225}$$ = $$x$$ 15 = x$$15$$ = $$x$$
-saknis(225) = x$$-\sqrt {225}$$ = $$x$$ -15 = x$$-15$$ = $$x$$
Parabolė kerta x ašį kai x = -15 ir x = 15. Todėl AB ilgis yra 15 + 15 = 30.
Atsakymas: 30
Sprendimas:
Kadangi pakylos ilgis yra 28, tai ji prasideda ties koordinate x = -14 ir baigiasi ties koordinate x = 14.
$$h = -0.1\cdot 14^{2}+22.5 = -0.1\cdot 196+22.5 = -19.6+22.5 = 2.9$$
Atsakymas: 2.9
Sprendimas:
∫(-15;15;- 0.1* x^2+22.5) =
∫(-15;15;- 0.1* x^2+22.5) = $$\int_{-15}^{15} (-0.1\cdot x^{2}+22.5)$$ =
∫(-15;15;- * x^2+22.5) = $$\int_{-15}^{15} (-\frac{1}{10}\cdot x^{2}+22.5)$$ = (- + 22.5* 15)-(- * (-15)^3+ 22.5* (-15)) = $$(-\frac{225}{2}+22.5\cdot 15)-(-\frac{1}{30}\cdot (-15)^{3}+22.5\cdot (-15))$$ = (- +337.5)-( * (-15)^3+ 22.5* (-15)) = $$(-\frac{225}{2}+337.5)-(\frac{1}{30}\cdot (-15)^{3}+22.5\cdot (-15))$$ = (- +337.5)-( - 22.5* (-15)) = $$(-\frac{225}{2}+337.5)-(\frac{225}{2}-22.5\cdot (-15))$$ = (- +337.5)-( -337.5) = $$(-\frac{225}{2}+337.5)-(\frac{225}{2}-337.5)$$ = - +337.5-( -337.5) = $$-\frac{225}{2}+337.5-(\frac{225}{2}-337.5)$$ = - +337.5- +337.5 = $$-\frac{225}{2}+337.5-\frac{225}{2}+337.5$$ = - - +337.5+337.5 = $$-\frac{225}{2}-\frac{225}{2}+337.5+337.5$$ = -225+337.5+337.5 = $$-225+337.5+337.5$$ =
-225+675 = $$-225+675$$ =
450$$450$$
Atsakymas: 450
21 uždavinys23 uždavinys