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3 years ago

Can anybody answer this?

Mathematics

2 answers:

kirza4 [7]3 years ago

7 0

Should be 180 my g, I’m pretty sure

lawyer [7]3 years ago

5 0

Step-by-step explanation:

${2}^{2} \cdot \: {3}^{2} \cdot5 \\ = 2\cdot \: 2\cdot \: 3\cdot \: 3\cdot5 \\ = 180$

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A sea bird is 28 meters above the surface of the ocean what is the elevation

Artyom0805 [142]

91.8 ft above sea level

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3 years ago

Jeany bought 12 postcards which was twice as many postcards as joe bought. Ruth bought 3 times as many postcards as joe. How man

nasty-shy [4]

Answer:36 postcards

Step-by-step explanation:

jenny bought 12 post cards

joe bought half of the postcards jenny bought = 12/2 = 6

ruth bought 3 times the postcards joe bought = 3 x 6 = 18

total postcards bought by all = 12 + 6 + 18 = 36 postcards

8 0

3 years ago

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A ball is thrown from a height of 70 meters with an initial downward velocity of 10/ms. The ball's height h (in meters) after t

dmitriy555 [2]

Answer:h=−5t2−10t+70

Step-by-step explanation:

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4 years ago

PLEASE HELP!!! THIS IS MY LAST QUESTION!!!! PLEASE!

sladkih [1.3K]

Answer:

2, -2, 0

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4 0

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