Let's Simplify X to equal 1 by dividing by 100. then we divide 420 by 100 to get 4.2, so y=4.2 when x=1, we then multiply by each by 11.
y = 46.2
x = 11
We have
Plug in
:
⇒
So we now have
Plug in
:
⇒
⇒
![b=\sqrt[3]{\frac{95}{4}}](https://tex.z-dn.net/?f=b%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B95%7D%7B4%7D%7D)
which is approximately 2.874
So we get
![y=4(\sqrt[3]{\frac{95}{4}})^{x}](https://tex.z-dn.net/?f=y%3D4%28%5Csqrt%5B3%5D%7B%5Cfrac%7B95%7D%7B4%7D%7D%29%5E%7Bx%7D)
or, in decimal form,
The distance between two points with the given coordinates in space is;
<u><em>Distance = 17 units</em></u>
We are given the coordinates;
(32, 12, 5)
(20, 3, 13)
- Formula for the distance between two points that have (x, y, z) coordinates is;
d = √((x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²)
From the equation given to us, we can see that;
x₁ = 32
x₂ = 20
y₁ = 12
y₂ = 3
z₁ = 5
z₂ = 13
- Using the <em>formula for the distance</em>, we have;
d = √((20 - 32)² + (3 - 12)² + (13 - 5)²)
d = √(144 + 81 + 64)
d = √289
d = 17
Thus, the <em>distance</em> between the two points is 17 units
Read more at; brainly.com/question/20974053
Step-by-step explanation:
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
2
+
6
+
4
=
0
2x^{2}+6x+4=0
2x2+6x+4=0
=
2
a={\color{#c92786}{2}}
a=2
=
6
b={\color{#e8710a}{6}}
b=6
=
4
c={\color{#129eaf}{4}}
c=4
=
−
6
±
6
2
−
4
⋅
2
⋅
4
√
2
⋅
2
x=\frac{-{\color{#e8710a}{6}} \pm \sqrt{{\color{#e8710a}{6}}^{2}-4 \cdot {\color{#c92786}{2}} \cdot {\color{#129eaf}{4}}}}{2 \cdot {\color{#c92786}{2}}}
x=2⋅2−6±62−4⋅2⋅4
brainliest and follow and thanks
