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3 years ago

Write the number shown on the calculator display in standard form.

Mathematics

1 answer:

frez [133]3 years ago

3 0

Answer:

2 Because visual specialization skills are important to success in so many career

fields, it is helpful to continue to build skill. What is one strategy you can use to

continue to improve your visualization skills?

Please someone help!!!

Step-by-step explanation:

2 Because visual specialization skills are important to success in so many career

fields, it is helpful to continue to build skill. What is one strategy you can use to

continue to improve your visualization skills?

Please someone help!!!

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HELP IM ON THE LAST ONE ILL GIVE BRAINLIEST

xxMikexx [17]

Answer:

this is hard what grade is this for

Step-by-step explanation:

8 0

3 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

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

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3 years ago

Please help me solve this problem

olga2289 [7]

1,025. Multiply 55 by 15, then add the $200 fee. I hope this helps :)

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3 years ago

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Slav-nsk [51]

Answer:

answers below

Step-by-step explanation:

1)5000

2)6450

3)50

4)30

5)1500

please make brainliest hope this helps

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3 years ago

Find the value of x in the figure below

Sonbull [250]

Answer:

Step-by-step explanation:

x = 1.0

how:

we know that the Hyp of the small triangle is $\sqrt{2}$

since we know that , then

$\sqrt{2}$ = 1^2 + 1^2 by Pythagoras theorem

so each side is 1 :)

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2 years ago