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igor_vitrenko [27]
3 years ago
10

Landon played football last week he played 22 minutes on Monday 19 minutes on Tuesday and 33 minutes on Wednesday on Thursday he

play twice as much as Monday if he played a total of 150 minutes last week how many minutes did he play on Friday
Mathematics
1 answer:
ladessa [460]3 years ago
4 0

Answer:

32

Step-by-step explanation:

M: 22

T:19

W:33

Th: 2m, if m=22, Th=44

22+19+33+44=118

150-118=32

So he played for 32 minutes on Friday

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deez  n u t s

Step-by-step explanation:

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3 years ago
The angle measurements in the diagram are represented by the following expressions.
nataly862011 [7]

Answer:

x=10

A = 84

Step-by-step explanation:

A and B are supplementary angles so they add to 180 degrees

A + B =180

2x+76 + 5x+34 = 180

Combine like terms

7x+ 110 = 180

7x+110-110 = 180-110

7x = 70

Divide by 7

x = 10

Now find A

A = 5x+34

A = 5*10 +34

  = 50+34

A = 84

7 0
3 years ago
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2x+3y=6<br> 3x-y=2<br> How many solutions does this have, one, none or infinite
sergey [27]
Y=-2/3x + 2
y=3x-2

one because the slopes are different

6 0
3 years ago
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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
3 years ago
Can someone please answer this m
sweet-ann [11.9K]

Answer:

5^-4

1 / 5^4

Step-by-step explanation:

5^3  / 5^7

We know a^b  / a^c  = a^(b-c)

5^(3-7)

5^-4

If you are not allowed to have negative exponents

We know a^-b = 1/a^b

1 / 5^4

7 0
3 years ago
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