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3 years ago

Can anyone help me with this question

Mathematics

1 answer:

GREYUIT [131]3 years ago

8 0

Reflect the triangle across the y-axis. Next translate it 2 units to the left. Then translate it 6 units down.

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Your family is traveling to Mexico on vacation. You have

ycow [4]

Answer:

5960 pesos

Step-by-step explanation:

We can write a proportion

298 pesos --- $25

x pesos --- $500

x=298*500/25 = 298*20 = 5960 pesos

8 0

3 years ago

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If 62% of 80 classmates like hip-hop, but 70% of 60 relatives like hip-hop, which one is more?

SIZIF [17.4K]

62 percent of 80 is more

Step-by-step explanation:

the 62 percent of 80 comes out to 49.6 people and the other one comes out to 42 people

i dont know how to explain how to do it

8 0

2 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

15 is what percent of<br> 600?<br> Pls show work

chubhunter [2.5K]

Answer:

Step-by-step explanation:

600(.15) = 90

8 0

2 years ago

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If the radius of a cone is 2 and the height is 4 what is the volume

laila [671]

Your answer: V≈16.76

4 0

3 years ago