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salantis [7]
3 years ago
8

Can someone help me solve this?

Mathematics
1 answer:
lozanna [386]3 years ago
6 0

Answer:

what uhhhhhh

Step-by-step explanation:

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Ian drinks 3 \text { pints}3 pints3, start text, space, p, i, n, t, s, end text of water during a long hike. He drinks 1\dfrac12
r-ruslan [8.4K]

Answer:

Total = 7\frac{1}{2} cups

Step-by-step explanation:

Given

Initially, he drank 3 pints of water

Then, 1\frac{1}{2} more cups

Required

How many cups did Ian drink altogether

This question can be solved by adding the size of water he drank during the hike and when he got home

i.e.

Total = 3\ pints + 1\frac{1}{2} cups

But the answer should be in cups;

So, number of pints has to be converted to number cups

Given that 1 pint = 2 cups;

3 pints will be 3 * 2 cups

3 pints = 6 cups;

So,

Total = 3\ pints + 1\frac{1}{2} cups

Total = 6\ cups + 1\frac{1}{2} cups

Total = 7\frac{1}{2} cups

Altogether, he drank 7\frac{1}{2} cups of water

5 0
3 years ago
Given f(x) = 5x - 6 and g(x) = x2 - 5x + 6, find the function (f + g)(x). Use the ^ key for the exponent. Enter your answer as t
sesenic [268]
The answer is x².

f(x) = <span>5x - 6
</span>g(x) = x²<span> - 5x + 6

(f + g)(x) = f(x) + g(x)
              = </span>5x - 6 + x² - 5x + 6
              = x² + 5x - 5x + 6 - 6
              = x² + 0 + 0
              = x²
6 0
3 years ago
HELP PLEASE :(( thank u
yaroslaw [1]

Answer:

the 3rd one hope that helped

Step-by-step explanation:

8 0
3 years ago
Use lagrange multipliers to find the point on the plane x â 2y + 3z = 6 that is closest to the point (0, 2, 4).
Arisa [49]
The distance between a point (x,y,z) on the given plane and the point (0, 2, 4) is

\sqrt{f(x,y,z)}=\sqrt{x^2+(y-2)^2+(z-4)^2}

but since \sqrt{f(x,y,z)} and f(x,y,z) share critical points, we can instead consider the problem of optimizing f(x,y,z) subject to x-2y+3z=6.

The Lagrangian is

L(x,y,z,\lambda)=x^2+(y-2)^2+(z-4)^2+\lambda(x-2y+3z-6)

with partial derivatives (set equal to 0)

L_x=2x+\lambda=0\implies x=-\dfrac\lambda2
L_y=2(y-2)-2\lambda=0\implies y=2+\lambda
L_z=2(z-4)+3\lambda=0\implies z=4-\dfrac{3\lambda}2
L_\lambda=x-2y+3z-6=0\implies x-2y+3z=6

Solve for \lambda:

x-2y+3z=-\dfrac\lambda2-2(2+\lambda)+3\left(4-\dfrac{3\lambda}2\right)=6
\implies2=7\lambda\implies\lambda=\dfrac27

which gives the critical point

x=-\dfrac17,y=\dfrac{16}7,z=\dfrac{25}7

We can confirm that this is a minimum by checking the Hessian matrix of f(x,y,z):

\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}

\mathbf H is positive definite (we see its determinant and the determinants of its leading principal minors are positive), which indicates that there is a minimum at this critical point.

At this point, we get a distance from (0, 2, 4) of

\sqrt{f\left(-\dfrac17,\dfrac{16}7,\dfrac{25}7\right)}=\sqrt{\dfrac27}
8 0
3 years ago
Which expression is equivalent to (2g^5)^3/(4h^2)^3
9966 [12]
\cfrac{(2g^5)^3}{(4h^2)^3} = \cfrac{2^3 \cdot g^{5\cdot3}}{4^3 \cdot h^{2\cdot3}} = \cfrac{8 g^{15}}{64 h^{6}} =\cfrac{g^{15}}{8h^{6}}

First option
7 0
3 years ago
Read 2 more answers
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