Answer:
1
Use the quadratic formula
=
−
±
2
−
4
√
2
x=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c}}}}{2{\color{#c92786}{a}}}
x=2a−b±b2−4ac
Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.
2
+
5
−
2
=
0
x^{2}+5x-2=0
x2+5x−2=0
=
1
a={\color{#c92786}{1}}
a=1
=
5
b={\color{#e8710a}{5}}
b=5
=
−
2
c={\color{#129eaf}{-2}}
c=−2
=
−
5
±
5
2
−
4
⋅
1
(
−
2
)
√
2
⋅
1
Step-by-step explanation:
this should help
Let u be the set of all words containing 7 letters(26 letters to the power 7)
5 students like cheesecake because if there is 20 students in all and 15 like chocolate and the rest like cheesecake you do 20-15=5
If 5 students like cheesecake out of 20 it would be 5/20 20*5=100 just multiply by 5 5*5=25%
25% of students like cheesecake
Step-by-step explanation:

In this case we have:
Δx = 3/n
b − a = 3
a = 1
b = 4
So the integral is:
∫₁⁴ √x dx
To evaluate the integral, we write the radical as an exponent.
∫₁⁴ x^½ dx
= ⅔ x^³/₂ + C |₁⁴
= (⅔ 4^³/₂ + C) − (⅔ 1^³/₂ + C)
= ⅔ (8) + C − ⅔ − C
= 14/3
If ∫₁⁴ f(x) dx = e⁴ − e, then:
∫₁⁴ (2f(x) − 1) dx
= 2 ∫₁⁴ f(x) dx − ∫₁⁴ dx
= 2 (e⁴ − e) − (x + C) |₁⁴
= 2e⁴ − 2e − 3
∫ sec²(x/k) dx
k ∫ 1/k sec²(x/k) dx
k tan(x/k) + C
Evaluating between x=0 and x=π/2:
k tan(π/(2k)) + C − (k tan(0) + C)
k tan(π/(2k))
Setting this equal to k:
k tan(π/(2k)) = k
tan(π/(2k)) = 1
π/(2k) = π/4
1/(2k) = 1/4
2k = 4
k = 2
Step-by-step explanation:
i think this pics will help u