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uranmaximum [27]

2 years ago

13

Kate has 12 1/2 pounds of chocolate. She gives each of her 5 friends x pounds each and has 3 1/3 pounds left over. How much did

she give each of her friends?

1 answer:

Alenkasestr [34]2 years ago

4 0

Let's write an equation and solve for x
12 1/2 = 5x + 3 1/3
rewrite the mixed fractions as improper fractions and with denominator 6
75/6 = 5x + 20/6
subtract 20/6 from both sides
55/6 = 5x
divide both sides by 5 to isolate x
11/6 = x
That is 1 and 5/6 pounds per each of the friends

To check:
<span>12 1/2 =?= 5(1 5/6) + 3 1/3
12 1/2 =?= 55/6 + 3 1/3
12 1/2 =?= 25/2
12 1/2 = 12 1/2

Hope this helps</span>

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LuckyWell [14K]

Answer:

Mid grade gas per gallon = $2.67

Regular gas per gallon = $2.37

Step-by-step explanation:

Let, x= mid grade gas, y=regular gas.

So, for 18 gallons of mid grade gas and 5 gallon of regular gas at $59.91, it can be expressed as,

$18x+5y=59.91$ ------------------(equation 1)

for 14 gallon of mid grade gas and 1 gallon of regular gas at 39.75, it can be expressed as,

$14x+y=39.75$ -------------------(equation 2)

$y=39.75-14x$ -------------------(equation 3)

Now substituting the value of y from (equation 3) in (equation 1) we get,

$18x+5y=59.91$

$18x+5(39.75-14x)=59.91$

$18x+198.75-70x=59.91$

$52x=198.75-59.91$

$52x=138.84$

$x=\frac{138.84}{52}$

$x=2.67$ -------------------------(equation 4)

Now substituting value of x from (equation 4) in (equation 3) we get,

$y=39.75-14x$

$y=39.75-(14\times2.67)$

$y=39.75-37.38$

$y=2.37$

Therefore price per gallon of mid grade gas is x = $2.67, and price per gallon of regular gas y = $2.37.

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2 years ago

Paraphin [41]

Answer: ok so for 5 and 125 it is 65 and 25

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3 years ago

How do you do these two questions?

Mice21 [21]

Answer:

a) d²y/dx² = ½ x + y − ½

b) Relative minimum

Step-by-step explanation:

a) Take the derivative with respect to x.

dy/dx = ½ x + y − 1

d²y/dx² = ½ + dy/dx

d²y/dx² = ½ + (½ x + y − 1)

d²y/dx² = ½ x + y − ½

b) At (0, 1), the first and second derivatives are:

dy/dx = ½ (0) + (1) − 1

dy/dx = 0

d²y/dx² = ½ (0) + (1) − ½

d²y/dx² = ½

The first derivative is 0, and the second derivative is positive (concave up). Therefore, the point is a relative minimum.

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3 years ago

Find the tangent plane to the given surface of f(x,y)=6- 6/5 x-y at the point (5, -1, 1). Make sure tat your final answer for th

HACTEHA [7]

Answer:

Required equation of tangent plane is $z=\frac{6}{5}(x-5y-11)$ .

Step-by-step explanation:

Given surface function is,

$f(x,y)=6-\frac{6}{5}(x-y)$

To find tangent plane at the point (5,-1,1).

We know equation of tangent plane at the point $(x_0,y_0,z_0)[/tex] is,

$z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)\hfill (1)$

So that,

$f(x_0,y_0)=6-\frac{6}{5}(5+1)=-\frac{6}{5}$

$f_x=-\frac{6}{5}y\implies f_x(5,-1,1)=\frac{6}{5}$

$f_y=-\frac{6}{5}x\implies f_y(5,-1,1)=-6$

Substitute all these values in (1) we get,

$z=\frac{6}{5}(x-5)-6(y+1)-\frac{6}{5}$

$\therefore z=\frac{6}{5}(x-5y-11)$

Which is the required euation of tangent plane.

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nevsk [136]

The answer is the second one I dont have that symbol on my keyboard but it’s the second answer on the multi choice
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2 years ago

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