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Vesnalui [34]
3 years ago
12

Hey please help me if you can I would appreciate it

Mathematics
2 answers:
PolarNik [594]3 years ago
6 0
This is a division problem.
You need to divide 51/2 by 3/2.

\dfrac{51}{2} \div  \dfrac{3}{2} =

= \dfrac{51}{2} \times  \dfrac{2}{3}

= \dfrac{51 \times 2}{2 \times 3}

= \dfrac{102}{6}

= 102 \div 6

= 17

Answer: Choice D.




MArishka [77]3 years ago
6 0
I think the answer is 38because the number of notebooks is 51/2 / 3/2
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Is 17/72 closer to 0, 0.5, or 1
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Step-by-step explanation:

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4 0
3 years ago
Given: ∆ABC, AB = BC, BM = MC<br> AC = 40, m∠BAC = 42º<br> Find: AM
Anna [14]

Answer:

The length of AM is 26.50 units.

Step-by-step explanation:

Given information:  AB = BC, BM = MC , AC = 40, ∠BAC = 42º.

Since two sides of triangle are equal, therefore the triangle ABC is an isosceles triangle.

The corresponding angles of congruents sides are always equal. So angle C is 42º.

According to the angle sum property the sum of interior angles is 180º.

\angle B=180-42-42=96

Law of Sine

\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}

\frac{\sinB}{AC}=\frac{\sin(C}{AB}

\frac{\sin(96)}{40}=\frac{\sin(42}{AB}

AB\sin(96)=40\sin(42)

AB=\frac{40\sin(42)}{\sin(96)}

AB=26.91

Therefore the length of AB and BC is 26.91.

Since M is midpoint of BC, so

BM=\frac{BC}{2}=\frac{26.91}{2}=13.455

Use Law of Cosine in triangle ABM to find the value of AM.

a^2=b^2+c^2-2bc\cos A

AM^2=AB^2+BM^2-2(AB)(BM)\cos (B)

AM^2=(26.91)^2+(13.455)^2-2(26.91)(13.455)\cos (96)

AM=26.50

Therefore the length of AM is 26.50 units.

6 0
3 years ago
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