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LekaFEV [45]
3 years ago
6

Claire get a breakfast to go at a local store. She eats 2 pieces of buttered toast and a glass of juice for breakfast every day.

Mathematics
2 answers:
Bond [772]3 years ago
4 0

Answer:

7(2B + J)

Step-by-step explanation:

Two slices of toast would come to 2B; 1 glass of juice to J.

Then Claire's daily expenditure for breakfast would be 2B + J.

For a 7-day week, this comes to   (Answer C)

Luda [366]3 years ago
3 0

Answer:

(C.) 7(J+2B)

Step-by-step explanation:

7 days times (1 glass of jiuce + 2 slice of bread)

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Which equation demonstrates the additive identity property?
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Answer:

Step-by-step explanation:

Solve for o in the equation (7 + 4) + (7 - 41) = 14©(7 + 4) + 0 = 7 + 41o(7 + 4)(1) = 7 + 41o(7 + 41) + ( - 7 - 41) = 0

We first need to simplify the expression removing parentheses

Simplify 41o(7 + 4): Distribute the 41o to each term in (7+4)

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Simplify 41o(7 + 41): Distribute the 41o to each term in (7+41)

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Our updated term to work with is (7 + 4) + (7 - 41) = 14©(7 + 4) + 0 = 7 + 287o + 164o(1) = 7 + 287o + 1681o + ( - 7 - 41) = 0

We first need to simplify the expression removing parentheses

Simplify 164o(1): Distribute the 164o to each term in (1)

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Our updated term to work with is (7 + 4) + (7 - 41) = 14©(7 + 4) + 0 = 7 + 287o + 164o = 7 + 287o + 1681o + ( - 7 - 41) = 0

Step 1: Group variables: We need to group our variables (7 and 14©(7. To do that, we subtract 14©(7 from both sides (7 - 14©(7 = 14©(7 - 14©(7

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0o 0 = 0 0 o =

5 0
3 years ago
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Given: KL ║ NM , LM = 45, m∠M = 50° KN ⊥ NM , NL ⊥ LM Find: KN and KL
Mice21 [21]

Answer:

KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08\\ \\KN=45\sin 50^{\circ}\approx 34.47

Step-by-step explanation:

Given:

KL ║ NM ,

LM = 45

m∠M = 50°

KN ⊥ NM  

NL ⊥ LM

Find: KN and KL

1. Consider triangle NLM. This is a right triangle, because NL ⊥ LM. In this triangle,

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\tan \angle M=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{NL}{LM}=\dfrac{NL}{45}\\ \\NL=45\tan 50^{\circ}

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2. Consider triangle NKL. This is a right triangle, because KN ⊥ NM . In this triangle,

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m\angle KNL=90^{\circ}-40^{\circ}=50^{\circ} (angles KNL and KLN are complementary).

So,

\sin \angle KNL=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{KL}{LN}=\dfrac{KL}{45\tan 50^{\circ}}\\ \\KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08

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\cos \angle KNL=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{KN}{LN}=\dfrac{KN}{45\tan 50^{\circ}}\\ \\KN=45\tan 50^{\circ}\cos 50^{\circ}=45\sin 50^{\circ}\approx 34.47

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