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3 years ago

9

A decimal that would be more than a half

1 answer:

Reika [66]3 years ago

4 0

Answer:

So basically anything higher than .5 would be x>.5 so it has to be greater than or equal to .51

Step-by-step explanation:

.51 .75 .89 .56 .78 .86 .61 it goes on and on as long as it is greater than .50

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You are trying to recreate flubber because you are obsessed with that movie and you thought it was the coolest. you take 70% sli

german

Let $x$ be the amount (gal) of the 40% slime solution you end up using. You want to end up with 120 gal of the 50% solution, so that you would use $120-x$ gal of the 70% solution.

You want the final solution to consist of 50% slime, or 60 gal of slime. Each gal of the 40% solution contributes 0.4 gal of slime, while each gal of the 70% solution contributes 0.7 gal of slime. This means

$0.4x+0.7(120-x)=0.5\cdot120=60$

$\implies0.4x+84-0.7x=60$

$\implies24=0.3x$

$\implies x=80$

so the answer is B.

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3 years ago

A store buys a jacket for $20.00 and then sells it to customers for 80% more than that. What is the selling price of the jacket?

Elis [28]

Your answer would be 36.

Hope this helped!

8 0

4 years ago

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You roll a number cube. What is the probability it will land on a number greater than 5?

olasank [31]

1 out of 6 chances

Explanation: there’s 6 sides of a cube if you mark them 1-6 there only one side that you can land on that is greater then 5 which is six also if you need it in percent it’s 16.6%

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2 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

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

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3 years ago

Help please first with the right answer will be marked brainlyest

andre [41]

I believe this is correct

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3 years ago

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