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2 years ago

13

X =Angle measure in the following picture solve for x

1 answer:

rjkz [21]2 years ago

8 0

Answer:

x=11

Step-by-step explanation:

3x+28+5x+52+12=180

8x+92=180

-92 on both sides

8x=88

/8 on both sides

x=11

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Mrs. Pascal has a candy jar that contains 5 Jolly Ranchers and 13 jawbreakers. Solve each of the following problems and explain

UkoKoshka [18]

Answer:

Step-by-step explanation:

The ratio is 5:13 and the total number of candy is 18. If the total amount of candy increased by 40x then both candy values would increase by 40x. There are 200 Jolly ranchers and 520 jawbreakers.

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2 years ago

at the car wash fundraiser jenna's team washed 14 cars in 45 minutes. if they keep up at that rate, how many cars can the team w

s2008m [1.1K]

You can write it as a proportion and then solve the proportion.
(2.5 hours is 150 minutes)
14 cars x cars
--------------- = -------------------
45 minutes 150 minutes

Then solve.

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3 years ago

9/10 + 9/100 + 9/1000 + 9/10000 + … =

Korvikt [17]

Answer: $0.9999$

Step-by-step explanation:

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2 years ago

will give brainliest Each exterior angle of a certain regular polygon is 20°, so each interior angle of that polygon must be ___

barxatty [35]

Answer:

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2 years ago

Read 2 more answers

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