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Minchanka [31]
3 years ago
13

If anyone has Insta could you please report my old account the username is ashlynnneill and the reason I want someone to report

it is so it would get taken down because I can’t login to it
Mathematics
2 answers:
Pavlova-9 [17]3 years ago
7 0

Answer:

hope

Step-by-step explanation:

hope

Elena-2011 [213]3 years ago
3 0

Answer:

ok i will!

Step-by-step explanation:

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Susan threw a softball 42 years on her first try and 51 1/3 yard on her second try. How much farther did she throw the softball
Andru [333]

Answer:

28 feet farther than 1st ball.

Step-by-step explanation:

We have been given that Susan threw a softball 42 yards on her first try and 51\frac{1}{3} yard on her second try.

To find second ball is how much farther from the 1st ball, we will subtract 42 yards from 51\frac{1}{3} yards.

\text{The second ball is farther from 1st ball}=51\frac{1}{3}\text{ yards}-42\text{ yards}

\text{The second ball is farther from 1st ball}=\frac{154}{3}\text{ yards}-42\text{ yards}

Let us have a common denominator.

\text{The second ball is farther from 1st ball}=\frac{154}{3}\text{ yards}-\frac{42*3}{3}\text{ yards}  

\text{The second ball is farther from 1st ball}=\frac{154}{3}\text{ yards}-\frac{126}{3}\text{ yards}

\text{The second ball is farther from 1st ball}=\frac{154-126}{3}\text{ yards}

\text{The second ball is farther from 1st ball}=\frac{28}{3}\text{ yards}

\text{1 yard}=3\text{ feet}

\frac{28}{3}\text{ yards}=\frac{28}{3}\times 3\text{ feet}

\frac{28}{3}\text{ yards}=28\text{ feet}

Therefore, Susan thrown the second ball 28 feet farther from the 1st ball.

7 0
3 years ago
You spin the spinner and flip a coin. Find the probability of the compound event.
ICE Princess25 [194]

Answer:

3

Step-by-step explanation:

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3 years ago
Match each value with its formula for ABC.
Ivahew [28]

For the triangle ABC use the sine theorem:

\dfrac{a}{\sin A}= \dfrac{b}{\sin B}= \dfrac{c}{\sin C}.

1. From \dfrac{a}{\sin A}= \dfrac{b}{\sin B} you have \dfrac{a}{b} \cdot \sin B=\sin A.

2. From \dfrac{b}{\sin B}= \dfrac{c}{\sin C} you have \dfrac{b}{c} \cdot \sin C=\sin B.

3. From \dfrac{a}{\sin A}= \dfrac{c}{\sin C} you have \dfrac{c}{a} \cdot \sin A=\sin C.

4. From \dfrac{a}{\sin A}= \dfrac{c}{\sin C} you have \dfrac{\sin A}{\sin C} \cdot c=a.

5. From \dfrac{a}{\sin A}= \dfrac{b}{\sin B} you have \dfrac{\sin B}{\sin A} \cdot a=b.

6. From \dfrac{b}{\sin B}= \dfrac{c}{\sin C} you have \dfrac{\sin C}{\sin B} \cdot b=c.

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