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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

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.

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3 years ago

You spin the spinner and flip a coin. Find the probability of the compound event.

ICE Princess25 [194]

Answer:

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Step-by-step explanation:

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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 $\dfrac{b}{c} \cdot \sin C=\sin B$ .

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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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