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3 years ago

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Mrs. Fansler gives 25 assignments while Ms. Taylor gives 20 assignments. What is the ratio of Mrs. Fansler’s assignments to Ms.

Taylor's assignments in simplest form? Write as a fraction using the / for the fraction bar (ex. 2/3 with no spaces).
WILL GIVE BRAINLIEST!!

1 answer:

Shkiper50 [21]3 years ago

7 0

Answer:

5/4?

Step-by-step explanation:

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Help me<br><br> Simplify: 7s + 7 - 4s - 3

attashe74 [19]

<h2>Answer:</h2><h2>3s + 4</h2><h2 /><h2>Hope this helps!!!</h2>

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2 years ago

Read 2 more answers

What is and equation line that passes through the point (8,-5) and is parallel to the line 5x+4y=24

GenaCL600 [577]

Answer:

5x + 4x = 20

Step-by-step explanation:

You have your x and y values and if you substitute them in you find yourself with 40 - 20 = 24 this doesn't make sense. So you get another answer which can be set to 20 and 40 - 20 = 20 therefore you you can make sure that your line intersects with your x and y values.

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3 years ago

Angle a and angle b are both supplementary angles. What must be true about the angles?

melamori03 [73]

Not sure if there're any answer choices of some sort,

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3 years ago

inna [77]

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, or</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, or</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, orad = 3d^2, or</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, orad = 3d^2, ora = 3d.</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, orad = 3d^2, ora = 3d.a = 2, d = 2/3.</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, orad = 3d^2, ora = 3d.a = 2, d = 2/3.The common ratio of the GP is 3/2. Answer.</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, orad = 3d^2, ora = 3d.a = 2, d = 2/3.The common ratio of the GP is 3/2. Answer.Check: (2 + 2/3), (2 + 3*2/3), (2 + 6*2/3)</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, orad = 3d^2, ora = 3d.a = 2, d = 2/3.The common ratio of the GP is 3/2. Answer.Check: (2 + 2/3), (2 + 3*2/3), (2 + 6*2/3)= 8/3, 4, 6</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, orad = 3d^2, ora = 3d.a = 2, d = 2/3.The common ratio of the GP is 3/2. Answer.Check: (2 + 2/3), (2 + 3*2/3), (2 + 6*2/3)= 8/3, 4, 6Common ratio: 4/(8/3) = 3/2 same as 6/4 = 3/2. Correct.</em>

Step-by-step explanation:

Am I right? just commet if I'm wrong

then <em>TANKYOU>^_^<!</em>

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2 years ago

David ate 1/4 of a chocolate pie and peter ate 2/5.What fraction of the pie still remains?

spin [16.1K]

Answer:

Step-by-step explanation:

Pie eaten by David and Peter = (1/4) + (2/5)

$= \frac{1*5}{4*5}+\frac{2*4}{5*4}\\\\=\frac{5}{20}+\frac{4}{20}\\\\=\frac{9}{20}$

Remaining pie = 11/20

7 0

3 years ago