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3 years ago

9

For homework, a student had to complete 15 problems total. If she finished 6 problems in class, what is the ratio of problems sh

e still needs to complete to problems that she's already finished?

1 answer:

Sedaia [141]3 years ago

3 0

6:15
which can be simplified to
2:5

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Erin built a wooden box to hold hay on her farm. The box is 3 m long, 1 m wide, and 1 m high. Hay costs $14 per cubic meter.

nydimaria [60]

Answer:

It will cost 42 to completely fill the box with hay.

have a good day <3

5 0

3 years ago

Use the Distributive Property to rewrite the algebraic expression. 7(6m + 7)

jeka57 [31]

Step-by-step explanation:

first, distribute 7 to 6m (this means 7 • 6m)

7 • 6m = 42m

then distribute 7 to 7 (which is 7 • 7)

7 • 7 = 49

your simplified expression is <u>42m + 49</u>. i hope this helps! have a great day <3

6 0

2 years ago

A rancher wishes to build a fence to enclose a 2250 square yard rectangular field. Along one side the fence is to be made of hea

Bess [88]

Answer:

The least cost of fencing for the rancher is $1200

Step-by-step explanation:

Let <em>x</em> be the width and <em>y </em>the length of the rectangular field.

Let <em>C </em>the total cost of the rectangular field.

The side made of heavy duty material of length of <em>x </em>costs 16 dollars a yard. The three sides not made of heavy duty material cost $4 per yard, their side lengths are <em>x, y, y</em>. Thus

$C=4x+4y+4y+16x\\C=20x+8y$

We know that the total area of rectangular field should be 2250 square yards,

$x\cdot y=2250$

We can say that $y=\frac{2250}{x}$

Substituting into the total cost of the rectangular field, we get

$C=20x+8(\frac{2250}{x})\\\\C=20x+\frac{18000}{x}$

We have to figure out where the function is increasing and decreasing. Differentiating,

$\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}$

Next, we find the critical points of the derivative

$20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30$

Because the length is always positive the only point we take is $x=30$ . We thus test the intervals $(0, 30)$ and $(30, \infty)$

$C'(20)=20-\frac{18000}{20^2} = -25 < 0\\\\C'(40)= 20-\frac{18000}{20^2} = 8.75 >0$

we see that total cost function is decreasing on $(0, 30)$ and increasing on $(30, \infty)$ . Therefore, the minimum is attained at $x=30$ , so the minimal cost is

$C(30)=20(30)+\frac{18000}{30}\\C(30)=1200$

The least cost of fencing for the rancher is $1200

Here’s the diagram:

3 0

2 years ago

Solve the system of equations - 8x + 3y = -17 and 3x - y = 7 by combining the<br> equlations.

NeTakaya

Given :

Two equations :

-8x + 3y = -17 ....1)

3x - y = 7 ....2)

To Find :

The solution of the system.

Solution :

Multiplying equation 2) by 3 and adding with equation 1), we get :

(-8x + 3y) + 3(3x - y) = -17 + 21

x = 4

Putting above value of x equation 2) we get :

12 - y = 7

y = 5

Therefore, solution of system is ( 4,5 ).

8 0

2 years ago

In ΔKLM, k = 700 cm, l = 460 cm and ∠M=78°. Find the length of m, to the nearest centimeter.

pashok25 [27]

Answer:

753

Step-by-step explanation:

m^2 =700^2 +460^-2 −2(700)(460)cos78

m^2 = 490000+211600-2(700)(460)(0.207912)

m^2 = 490000+211600-133895.12889

m^2 = 567704.87111

m=753.462=753

6 0

2 years ago