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2 years ago

PLS HELP I NEED HELP ASAP

Mathematics

2 answers:

melisa1 [442]2 years ago

8 0

960 cubic centimeters

monitta2 years ago

6 0

Answer:

960 cubic centimeters

Step-by-step explanation:

Base area is area of triangle: b*h*1/2

= 12*8*1/2 = 48

Base area times height is 48*20 = 960

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melisa1 [442]

Answer is 34 ft squared

3 0

3 years ago

I need help!? Someone help me

hammer [34]

Answer 3 his car is slowing down

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

What is the value of X?

zhenek [66]

Tough sorry wish I could help u can’t too hard but I think it is equivalent to the other triangle so I think it’s 31? I THINK

7 0

3 years ago

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In the adjoining figure, the parallel lines are marked by arrow lines. The value of angle 1 is:

dlinn [17]

Answer:

20°

Step-by-step explanation:

40°, 70° and 90° are the measures of the three angles of the quadrilateral.

Measure of fourth angle of the Quadrilateral

= 360° - (40° + 70° + 90°)

= 360° - 200°

= 160°

Measure of angle 1 will be equal to the measure of the linear pair angle of 160° as they are corresponding angles.

Thus,

$m\angle 1 = 180\degree- 160\degree$

$\implies\huge\purple{\boxed{ m\angle 1 = 20\degree}}$

Alternate method:

$m\angle 1 = 180\degree- [360\degree-(40\degree+70\degree+90\degree)]$

$\implies m\angle 1 = 180\degree- [360\degree-200\degree]$

$\implies m\angle 1 = 180\degree- 160\degree$

$\implies m\angle 1 = 20\degree$

6 0

2 years ago

Read 2 more answers