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2 years ago

Somebody please help !

Mathematics

2 answers:

vovangra [49]2 years ago

7 0

Answer:

Translation

Step-by-step explanation:

NemiM [27]2 years ago

3 0

Answer:

7 for your x axis and -3 for Y

So your are going from G to H, thats important.

You go to the right seven times and down (hence -3) three times

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Solve the equation <br><br> 10(0.4+0.5g)=4g<br><br> g=?

Ivenika [448]

Step 1: 10(0.4+0.5g)=4g
Step 2: 4 + 5g = 4g
Step 3: 4 = -1g
Step 4: -4 = g :)

8 0

3 years ago

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An amusement park sold 27 child tickets. The other 23 tickets it sold were adult tickets. What is the ratio of the number of adu

amm1812

Answer:

23/50

Step-by-step explanation:

27 + 23 = 50 (Implying that 50 is the total amount of tickets).

23 of them were sold to adults, so basically 23/50.

Remember a ratio can be written in 3 forms, so it can be either of these:

23:50

23/50

23 to 50

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2 years ago

Four of the angle of a Pentagon are 70°,90°,100° and 145°. What is the side of a fifth angle?

dezoksy [38]

Answer: 135°

Step-by-step explanation:

The sum of the angles in a pentagon is equal to 540°, and the sum of the first four angles is 405°.

70 + 90 + 100 + 145 = 405

540 - 405 = 135

6 0

3 years ago

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

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WITCHER [35]

What is the slope? We need the slope to find y

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3 years ago