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algol [13]
3 years ago
7

Lala went to the store to buy snacks for her grandmother. She bought 1.4 pounds of cashews that costs $8.90 per pound. She also

bought 0.7 pounds of walnuts that cost $8.10 per pound. Did Lala spend more on cashews or walnuts? How much more?
Mathematics
2 answers:
Effectus [21]3 years ago
8 0

Answer:

Step-by-step explanation:

She bought 1.4 pounds of cashews that costs $8.90 per pound. This means that the total amount that she paid for 1.4 pounds of cashew is

8.9 × 1.4 = $12.46

She also bought 0.7 pounds of walnuts that cost $8.10 per pound. This means that the total amount that she paid for 0.7 pounds of walnuts is

0.7 × 8.1 = $5.67

Since $12.46 is greater than $5.67, it means that she spent more on Cashews than walnuts. The amount that she spent more on cashews is

12.46 - 5.67 = $6.79

r-ruslan [8.4K]3 years ago
3 0
Yes.she spent $6.79cent on cashews
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Answer:

1.

Tan 45=√2/n

n=√2

again

sin 45=√2/m

m=2

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<u>2</u><u>.</u>

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x=3/√2=3√2/2

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<u>3</u><u>.</u>

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4.

b=4 base sides of isosceles triangle

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ans:<u>a</u><u>=</u><u>4</u><u>√</u><u>2</u><u> </u><u>and</u><u> </u><u>b</u><u>=</u><u>4</u>

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3 years ago
4) The path of a satellite orbiting the earth causes it to pass directly over two
Naily [24]

Answers:

  • Satellite is approximately <u>2446.43 km</u> from station A.
  • Satellite is approximately <u>2441.61 km</u> above the ground.

=========================================================

Explanation:

I'm assuming tracking stations A and B are at the same elevation and are on flat ground. In reality, this is likely not the case; however, for the sake of simplicity, we'll assume this is the case.

The diagram is shown below. Points A and B describe the two stations, while point C is the satellite's location. Point D is on the ground directly below the satellite. We have these lengths

  • AB = 60 km
  • AD = x
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Focusing on triangle ACD, we can apply the tangent rule to isolate h.

tan(angle) = opposite/adjacent

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tan(86.4) = h/x

x*tan(86.4) = h

h = x*tan(86.4)

We'll use this later in the substitution below.

--------------------

Now move onto triangle BCD. For the reference angle B = 85, we can use the tangent rule to say

tan(angle) = opposite/adjacent

tan(B) = CD/DB

tan(B) = CD/(DA+AB)

tan(85) = h/(x+60)

tan(85)*(x+60) = h

tan(85)*(x+60) = x*tan(86.4) .............  apply substitution; isolate x

x*tan(85)+60*tan(85) = x*tan(86.4)

60*tan(85) = x*tan(86.4)-x*tan(85)

60*tan(85) = x*(tan(86.4)-tan(85))

x*(tan(86.4)-tan(85)) = 60*tan(85)

x = 60*tan(85)/(tan(86.4)-tan(85))

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--------------------

We'll use this approximate x value to find h

h = x*tan(86.4)

h = 153.612786190499*tan(86.4)

h = 2441.60531869599

h = 2441.61 km  is how high the satellite is above the ground.

Return to triangle ACD. We'll use the cosine rule to determine the length of the hypotenuse AC

cos(angle) = adjacent/hypotenuse

cos(A) = AD/AC

cos(86.4) = x/AC

cos(86.4) = 153.612786190499/AC

AC*cos(86.4) = 153.612786190499

AC = 153.612786190499/cos(86.4)

AC = 2446.43279498247

AC = 2446.43 km is the distance from the satellite to station A.

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