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photoshop1234 [79]
3 years ago
13

Dana and her 4 friends are going on a camping trip. Dana wants to see how much fruit she can share so everyone will get the same

amount of fruit. There are 5 apples, 10 oranges, 5 strawberries and 10 grapes. How much will each person get?
Mathematics
1 answer:
kupik [55]3 years ago
6 0
I agree that the answer is 6. 30÷5=6
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If you roll a six-sided die, what is the probability that you will roll a 3?
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the probability is 1/6, or 16.6%

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Analyze the diagram below and answer the question that follows.
babunello [35]

Shaded part means the yellow part

  • The two sets are S and Y.
  • The centre intersection part is A\cap
  • All the areas occupied by both sets is AUB .

Rest part :-

\\ \sf\longmapsto E-AUB=(AUB)^c

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Plz help me I just started 8th grade.
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Answer:

I think this is slope. So for each point you go up 3 units. All of the point should connect into a straight line.

Step-by-step explanation:

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3 years ago
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Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o
Brrunno [24]

Answer:

AB = \sqrt{a^2 + b^2-2abCos\ C}

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

Sin\ C = \frac{h}{a}

Make h the subject:

h = aSin\ C

Also, in BOC (Using Pythagoras)

a^2 = h^2 + x^2

Make x^2 the subject

x^2 = a^2 - h^2

Substitute aSin\ C for h

x^2 = a^2 - h^2 becomes

x^2 = a^2 - (aSin\ C)^2

x^2 = a^2 - a^2Sin^2\ C

Factorize

x^2 = a^2 (1 - Sin^2\ C)

In trigonometry:

Cos^2C = 1-Sin^2C

So, we have that:

x^2 = a^2 Cos^2\ C

Take square roots of both sides

x= aCos\ C

In triangle BOA, applying Pythagoras theorem, we have that:

AB^2 = h^2 + (b-x)^2

Open bracket

AB^2 = h^2 + b^2-2bx+x^2

Substitute x= aCos\ C and h = aSin\ C in AB^2 = h^2 + b^2-2bx+x^2

AB^2 = h^2 + b^2-2bx+x^2

AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2

Open Bracket

AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C

Reorder

AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C

Factorize:

AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C

In trigonometry:

Sin^2C + Cos^2 = 1

So, we have that:

AB^2 = a^2 * 1 + b^2-2abCos\ C

AB^2 = a^2 + b^2-2abCos\ C

Take square roots of both sides

AB = \sqrt{a^2 + b^2-2abCos\ C}

6 0
3 years ago
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