You would subtract 1 1/2 from 7 1/2 so the answer would be 6 hope I helped
The area of the part of the plane 3x 2y z = 6 that lies in the first octant is mathematically given as
A=3 √(4) units ^2
<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>
Generally, the equation for is mathematically given as
The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:
The partial derivatives of a function are f x and f y.
When these numbers are plugged into equation (1) and the integrals are given bounds, we get:
![&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\](https://tex.z-dn.net/?f=%26%3D%5Cint_%7B0%7D%5E%7B2%7D%20%5Cint_%7B0%7D%5E%7B3-%5Cfrac%7B3%7D%7B2%7D%20x%7D%20%5Csqrt%7B%28-3%29%5E%7B2%7D%2B%28-2%29%5E2%2B1dxdy%7D%20%5C%5C%5C%5C%26%3D%5Cint_%7B0%7D%5E%7B2%7D%20%5Cint_%7B0%7D%5E%7B3-%5Cfrac%7B3%7D%7B2%7D%20x%7D%20%5Csqrt%7B14%7D%20d%20x%20d%20y%20%5C%5C%5C%5C%26%3D%5Csqrt%7B14%7D%20%5Cint_%7B0%7D%5E%7B2%7D%5By%5D_%7B0%7D%5E%7B3-%5Cfrac%7B3%7D%7B2%7D%20x%7D%20d%20x%20d%20y%20%5C%5C%5C%5C%26%3D%5Csqrt%7B14%7D%20%5Cint_%7B0%7D%5E%7B2%7D%5Cleft%5B3-%5Cfrac%7B3%7D%7B2%7D%20x%5Cright%5D%20d%20x%20%5C%5C%5C%5C)
![&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}](https://tex.z-dn.net/?f=%26%3D%5Csqrt%7B14%7D%5Cleft%5B3%20x-%5Cfrac%7B3%7D%7B2%7D%20%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%20%5Ccdot%20x%5E%7B2%7D%5Cright%5D_%7B0%7D%5E%7B2%7D%20%5C%5C%5C%5C%26%3D%5Csqrt%7B14%7D%5Cleft%5B3-%5Cfrac%7B3%7D%7B2%7D%20%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%20%5Ccdot%20x%5E%7B2%7D%5Cright%5D_%7B0%7D%5E%7B2%7D%20%5C%5C%5C%5C%26%3D%5Csqrt%7B14%7D%5Cleft%5B3.2-%5Cfrac%7B3%7D%7B2%7D%20%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%20%5Ccdot%203%5E%7B2%7D%5Cright%5D%20%5C%5C%5C%5C%26%3D3%20%5Csqrt%7B14%7D%20%5Ctext%20%7B%20units%20%7D%7B%20%7D%5E%7B2%7D)
In conclusion, the area is
A=3 √4 units ^2
Read more about the plane
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1 conclusion you can make is that p is the bisector of the line. And another is that QP and PR are congruent.
Sorry if not right :/
Answer: 1.5%
Step-by-step explanation:
The Pure Interest Rate refers to the rate in a theoretical market where there is no risk as economic conditions can be predicted with certainty.
Sadly, uncertainty will always exist but there are rates that are close enough to the Pure interest rate such as US Government Security rates.
The United States has never defaulted on a debt payment in the modern era and as such is known to offer the least risky rates in the world. The rate on US Securities can therefore be considered as close as possible to the pure interest rate.
Given the rates in the question, the Pure Interest rate would be the 20-year Treasury bond rate of 1.5%.
Answer:
They have the same x-value
f(x) has the greater minimum
Step-by-step explanation:
To find the vertex of a second degree equation, in this case the minimum value, we can use the following equation:
x = -b / 2a
Remember that a second degree equation has the following form:
ax^2 + bx + c
so a = 1, b = -8 and c = 7. Now you have to substitute in the previous equation
x = - (-8) / 2(1)
x = 8 / 2
x = 4
This means that the two functions have the same x-value.
The y value of f(x) would be
f(4) = (4)^2 - 8(4) + 7
f(4) = 16 - 32 + 7
f(4) = -9
So the vertex, or minimun value of f(x) would be at the point (4, -9).
The vertex, or minimun value of g(x) is at the point (4, -4).
So f(x) has a minimum value of -9 and g(x) a minimum value of -4.
