Answer:
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.
5
2
−
4
1
+
8
=
0
5x^{2}-41x+8=0
5x2−41x+8=0
=
5
a={\color{#c92786}{5}}
a=5
=
−
4
1
b={\color{#e8710a}{-41}}
b=−41
=
8
c={\color{#129eaf}{8}}
c=8
=
−
(
−
4
1
)
±
(
−
4
1
)
2
−
4
⋅
5
⋅
8
√
2
⋅
5
2
Simplify
3
Separate the equations
4
Solve
Solution
=
8
=
1
5
300 ml ... 200 ml
75 ml ... x = ?
If you would like to know what is the final volume of a gas, you can calculate this using the following steps:
300 * x = 200 * 75 /300
x = 200 * 75 / 300
x = 50.0 ml
The correct result would be A. 50.0 ml.
Answer:
- the given dimension was used as the radius
- 5.57 m³
Step-by-step explanation:
The volume of a sphere can be found using the formula ...
V = 4/3πr³ . . . . . where r is the radius
__
The figure points to a diameter line and indicates 2.2 m. The arrowhead is in the middle of a radius line, making it easy to interpret the dimension as the radius of the sphere.
If 2.2 m is used as the radius, the volume is computed to be ...
V = 4/3π(2.2 m)³ ≈ 44.58 m³
This agrees with your friend's volume, suggesting the diameter was used in place of the radius in the computation.
__
The correct volume, using 2.2 m as the diameter, is ...
V = 4/3π(1.1 m)³ ≈ 5.57 m³
Answer:
d. HL CBCTC
Step-by-step explanation:
The hypotenuse and one leg are congruent so the reason for congruency is HL.
