Answer:
14/1.8 = x/4
Step-by-step explanation:
Using similar triangles,
height of man/length of man's shadow = height of tree/length of tree's shadow
1.8/4 = x/14
x/14 = 1.8/4
If we cross-multiply, we have
1.8 × 14 = x × 4
dividing both sides by 4 and 1.8,we have
14/4 = x/1.8
x/1.8 = 14/4
So, the two expressions we have are
x/14 = 1.8/4 and x/1.8 = 14/4.
So, the answer is 14/1.8 = x/4 since the product 1.8 × 14 = x × 4 cannot be expressed in the given ratio.
Answer:
30 will be in each bag
5 chocolate covered raisins will be left
Step-by-step explanation:
185 divided by 6=30
30*6=180
185-180=5
Step-by-step explanation:
P(t) = 12,000 (2)^(-t/15)
9,000 = 12,000 (2)^(-t/15)
0.75 = 2^(-t/15)
ln(0.75) = ln(2^(-t/15))
ln(0.75) = (-t/15) ln(2)
-15 ln(0.75) / ln(2) = t
t = 6.23
You might have made an error the first time you solved for x. I got x = -0.5.
When you have your log base 4, the way you cancel that out is by making 4 the base on both sides, so you get 4^(log4) to reduce to 1, and you're left with:
2x + 3 = 4^(1/2) ... Simplify
2x + 3 = 2
2x = -1
x = -1/2
If you plug that back in, everything checks out. Maybe double check your use of logarithm/exponent properties?
<span>For the plane, we have z = 5x + 9y
For the region, we first find its boundary curves' points of intersection.
x = x^4 ==> x = 0, 1.
Since x > x^4 for y in [0, 1],
The volume of the solid equals
![\int\limits^1_0 { \int\limits_{x^4}^x {(5x+9y)} \, dy } \, dx = \int\limits^1_0 {\left[5xy+ \frac{9}{2} y^2\right]_{x^4}^{x}} \, dx \\ \\ =\int\limits^1_0 {\left[\left(5x(x)+ \frac{9}{2} (x)^2\right)-\left(5x(x^4)+ \frac{9}{2} (x^4)^2\right)\right]} \, dx \\ \\ =\int\limits^1_0 {\left(5x^2+ \frac{9}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx =\int\limits^1_0 {\left( \frac{19}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx \\ \\ =\left[ \frac{19}{6} x^3- \frac{5}{6} x^6- \frac{1}{2} x^9\right]^1_0](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E1_0%20%7B%20%5Cint%5Climits_%7Bx%5E4%7D%5Ex%20%7B%285x%2B9y%29%7D%20%5C%2C%20dy%20%7D%20%5C%2C%20dx%20%3D%20%5Cint%5Climits%5E1_0%20%7B%5Cleft%5B5xy%2B%20%5Cfrac%7B9%7D%7B2%7D%20y%5E2%5Cright%5D_%7Bx%5E4%7D%5E%7Bx%7D%7D%20%5C%2C%20dx%20%20%5C%5C%20%20%5C%5C%20%3D%5Cint%5Climits%5E1_0%20%7B%5Cleft%5B%5Cleft%285x%28x%29%2B%20%5Cfrac%7B9%7D%7B2%7D%20%28x%29%5E2%5Cright%29-%5Cleft%285x%28x%5E4%29%2B%20%5Cfrac%7B9%7D%7B2%7D%20%28x%5E4%29%5E2%5Cright%29%5Cright%5D%7D%20%5C%2C%20dx%20%20%5C%5C%20%20%5C%5C%20%3D%5Cint%5Climits%5E1_0%20%7B%5Cleft%285x%5E2%2B%20%5Cfrac%7B9%7D%7B2%7D%20x%5E2-5x%5E5-%20%5Cfrac%7B9%7D%7B2%7D%20x%5E8%5Cright%29%7D%20%5C%2C%20dx%20%3D%5Cint%5Climits%5E1_0%20%7B%5Cleft%28%20%5Cfrac%7B19%7D%7B2%7D%20x%5E2-5x%5E5-%20%5Cfrac%7B9%7D%7B2%7D%20x%5E8%5Cright%29%7D%20%5C%2C%20dx%20%5C%5C%20%20%5C%5C%20%3D%5Cleft%5B%20%5Cfrac%7B19%7D%7B6%7D%20x%5E3-%20%5Cfrac%7B5%7D%7B6%7D%20x%5E6-%20%5Cfrac%7B1%7D%7B2%7D%20x%5E9%5Cright%5D%5E1_0)
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