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4 years ago

Write 6/9 in simplest form

1 answer:

5 0

First find numbers that the denominator and the numerator can both be divided with. Such as 3, now 6/3=2 and 9/3=3. So basically it is 2/3. 6/9 simplify to 2/3.

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Circle has an area of 841π cm2. What is the diameter?

Maru [420]

The diameter is approximately 32.72 cm.

7 0

3 years ago

You are making fruit baskets using 54 apples, 36 oranges, and 73 bananas. Explain why you cannot make identical fruit baskets wi

liq [111]

The bananas are not an equal number so you would not have the same amount unless you take one out.

7 0

3 years ago

Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Flura [38]

Answer:

x=-3

Step-by-step explanation:

-4(-2x+5)-5x+5=-24

8x-20-5x+5=-24

3x-15=-24

3x=-9

x=-3

I hope that this helped you!

4 0

3 years ago

Read 2 more answers

Priya’s family exchanged 250 dollars for 4,250 pesos. Priya bought a sweater for 510 pesos. How many dollars did the sweater cos

I am Lyosha [343]

Answer:

$30

Step-by-step explanation:

Dollars : Pesos

250 : 4250

X : 510

250/4250 = X/510

X = 30

5 0

3 years ago

Read 2 more answers

Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1

sleet_krkn [62]

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}$

From Left side:

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)$

$\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}$

NOTE: sin²θ + cos²θ = 1

$\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}$

$\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}$

$\dfrac{2sec^2\theta}{tan^2\theta-1}$

Left side = Right side <em>so proof is complete</em>

8 0

3 years ago

Read 2 more answers