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guapka [62]
2 years ago
13

a landscaper is supposed to plant pine trees, oak trees, and elm trees in a ratio of 48:12:120. At this time, the landscaper onl

y has 12 pine trees to plant. in order to keep up the same ratio, how many oak and elm trees will he need to plant? ​
Mathematics
2 answers:
iogann1982 [59]2 years ago
5 0

Answer:

Step-by-step explanation:

Coniferous trees keep their needles all year with the exception of tamarack. They are good trees to plant for privacy and wind breaks or shelterbelts

Pines include native white, red, and jack. They have long needles.

Spruces, black and white, and firs. They have short needles. They are important sources of wood fiber in northern Minnesota, and are excellent choices for windbreaks or shelterbelts.

Cedars include white or red. Cedars have scaled needles. Smaller than pines and spruces, cedars can provide wildlife cover and food.

Neko [114]2 years ago
3 0

Answer:

Step-by-step explanation:

Coniferous trees keep their needles all year with the exception of tamarack. They are good trees to plant for privacy and wind breaks or shelterbelts

Pines include native white, red, and jack. They have long needles.

Spruces, black and white, and firs. They have short needles. They are important sources of wood fiber in northern Minnesota, and are excellent choices for windbreaks or shelterbelts.

Cedars include white or red. Cedars have scaled needles. Smaller than pines and spruces, cedars can provide wildlife cover and food.

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969

Step-by-step explanation:

173 +  212 + 185 +  197 + 202  = 969

7 0
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The answer is 5 cm for the circumference of the smaller square
6 0
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Graph the line with slope -2/3<br> passing through the point (-4,5).
ELEN [110]

Answer:

graph this equation

y = -2/3x 7/3

Step-by-step explanation:

y-5=-2/3(x--4)

y-5 = -2/3x - 8/3

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4 0
2 years ago
What is the multiplicative inverse of 5 in z11, z12, and z13? you can do a trial-and-error search using a calculator or a pc?
grin007 [14]

A multiplicative inverse of an integer a is an integer x such that the product ax is congruent to 1 with respect to the modulus m.  

1. Z_{11}=\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4},\overline{5},\overline{6},\overline{7},\overline{8},\overline{9},\overline{10}\}.

Check:

  • 5\cdot 0=\overline{0};
  • 5\cdot 1=\overline{5};
  • 5\cdot 2=\overline{10};
  • 5\cdot 3=15=\overline{4};
  • 5\cdot 4=20=\overline{9};
  • 5\cdot 5=25=\overline{3};
  • 5\cdot 6=30=\overline{8};
  • 5\cdot 7=35=\overline{2};
  • 5\cdot 8=40=\overline{7};
  • 5\cdot 9=45=\overline{1};
  • 5\cdot 10=50=\overline{6}.

The multiplicative inverse of 5 in Z_{11} is 9.

2.   Z_{12}=\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4},\overline{5},\overline{6},\overline{7},\overline{8},\overline{9},\overline{10},\overline{11}\}.

Check:

  • 5\cdot 0=\overline{0};
  • 5\cdot 1=\overline{5};
  • 5\cdot 2=\overline{10};
  • 5\cdot 3=15=\overline{3};
  • 5\cdot 4=20=\overline{8};
  • 5\cdot 5=25=\overline{1};
  • 5\cdot 6=30=\overline{6};
  • 5\cdot 7=35=\overline{11};
  • 5\cdot 8=40=\overline{4};
  • 5\cdot 9=45=\overline{9};
  • 5\cdot 10=50=\overline{2};
  • 5\cdot 11=55=\overline{7}.

The multiplicative inverse of 5 in Z_{12} is 5.

3.  Z_{13}=\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4},\overline{5},\overline{6},\overline{7},\overline{8},\overline{9},\overline{10},\overline{11},\overline{12}\}.

Check:

  • 5\cdot 0=\overline{0};
  • 5\cdot 1=\overline{5};
  • 5\cdot 2=\overline{10};
  • 5\cdot 3=15=\overline{2};
  • 5\cdot 4=20=\overline{7};
  • 5\cdot 5=25=\overline{12};
  • 5\cdot 6=30=\overline{4};
  • 5\cdot 7=35=\overline{9};
  • 5\cdot 8=40=\overline{1};
  • 5\cdot 9=45=\overline{6};
  • 5\cdot 10=50=\overline{11};
  • 5\cdot 11=55=\overline{3};
  • 5\cdot 12=60=\overline{8}.

The multiplicative inverse of 5 in Z_{13} is 8.

8 0
3 years ago
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Paraphin [41]
This gives you three simultaneous equations:

6 = a + c
7 = 4a + c
1 = c
 
<u>c = 1
</u>
<u /><u />
If c =1,

6 = a + 1
<u>a = 5
</u>
<u /><u />
This doesn't work in the second equation, so the quadratic that goes through these points is not in the form y = ax^2 + bx + c
Was there supposed to be a b in the equation?




5 0
3 years ago
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