-85/6 or -14.17
first convert into improper fraction then multiply out
Answer:
x ≈ 58.9
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Geometry</u>
- [Right Triangles Only] tan∅ = opposite over adjacent
Step-by-step explanation:
<u>Step 1: Define</u>
We are given a right triangle. We can use trig to find the missing side length.
<u>Step 2: Identify Variables</u>
<em>POV from the angle measure</em>
Angle = 23°
Opposite Leg = 25
Adjacent Leg = <em>x</em>
<em />
<u>Step 3: Solve for </u><em><u>x</u></em>
- Substitution [tangent]: tan23° = 25/x
- Multiply <em>x</em> on both sides: xtan23° = 25
- Isolate <em>x</em>: x = 25/tan23°
- Evaluate: x = 58.8963
- Round: x ≈ 58.9
Answer:
40 % of the snacks were carrot sticks.
Step-by-step explanation:
Total number of carrot sticks in lunch = 18
Total number of apple slices in lunch = 27
So, Total Snacks in the Lunch = Carrot Sticks + Apple Slices
= 18 + 27 = 45
Now,
=
⇒Percentage of Carrot Sticks = 40%
Hence, 40 % of the snacks were carrot sticks.
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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