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balandron [24]
3 years ago
10

Sandra throws an object straight up into the air with initial velocity of 38 ft per square from platform that is 30 ft above the

ground how long will it take for object to hit the ground
Mathematics
1 answer:
expeople1 [14]3 years ago
7 0

Answer:

3 seconds.

Step-by-step explanation:

We have been given that Sandra throws an object straight up into the air with initial velocity of 38 ft per second from platform that is 30 ft above the ground.

To find the time it will take for object to hit the ground, we will use formula:  

h(t)=-16t^2+v_0t+h_0, where, v_0 represents initial velocity and h_0 represents initial height.

Upon substituting our given initial velocity and initial height, we will get:

h(t)=-16t^2+38t+30

We know that object will hit the ground, when height will be 0, so we will equate object's height equal to 0 and solve for t as:

-16t^2+38t+30=0

Divide by 2:

-8t^2+19t+15=0

Upon using quadratic formula, we will get:

t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-19\pm\sqrt{19^2-4(-8)(15)}}{2(-8)}

t=\frac{-19\pm\sqrt{361+480}}{-16}=\frac{-19\pm\sqrt{841}}{-16}

t=\frac{-19\pm29}{-16}

t=\frac{-19-29}{-16},t=\frac{-19+29}{-16}

t=\frac{-48}{-16},t=\frac{10}{-16}

t=3,t=-\frac{5}{8}

Since time cannot be negative, therefore, it will take 3 seconds for object to hit the ground.

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3 0
3 years ago
What is the area of a square with lengths of 7 inches
ludmilkaskok [199]

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

                  Hello!

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

❖ The area of a square with lengths of 7 inches is 49 inches.

In a square, all sides are equal so if one side is 7 inches, the rest of the sides are 7 inches. You multiply 7 x 7 because you only need 2 sides to find the area because the area of a square is l x w. (If another sides length is stated, that is the length that you will use.

~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ

3 0
3 years ago
Read 2 more answers
14 more than the prouduct of the three and nine
Varvara68 [4.7K]
41
3 times 9 is 27
27+14=41
5 0
3 years ago
Read 2 more answers
A complex shape with measurements 3 feet, 11 feet, 3 feet, 4 feet, 6 feet, 3 feet, 6 feet, 4 feet. What is the perimeter of the
Anna71 [15]

Answer:

Solution given:

perimeter=sum of all side =3 feet+ 11 feet+ 3 feet+ 4 feet+ 6 feet+ 3 feet+ 6 feet+4 feet

=40feet

is your answer

6 0
3 years ago
Jerome has 1/4 of the groups video games at his house. Mario has 2/5 of the groups video games at his house. What fraction of th
romanna [79]
13/20 of the group's video games are at Jerome or Mario's house.

Jerome has 1/4 of the group's video games.
Mario has 2/5 of the group's video games.

The question asks what fraction of the video games are at either of their houses. Therefore, we need to add the fractions together to find out the total fraction of the video games at both of their houses.

1/4  +  2/5
We can't add these together because they have different denominators. We'll need to find the least common denominator of 4 and 5 before we can add them. If you can't think of a number that the two denominators can both multiply to get, an easy way to find the common denominator is to multiply the numbers together.

4 x 5 = 20
Therefore, our common denominator is 20. Now, we need to convert the fractions so that they can have a denominator of 20.

1/4 = ?/20
In order to find the numerator, ask yourself what we multiplied the denominator by to get the new denominator. In this case, we multiplied 4 by 5 to get 20. Therefore, we need to multiply 1 by 5 to get an equal fraction.

1 x 5 = 5
1/4 = 5/20  
Our fraction is 5/20.

2/5 = ?/20
Now, let's do the same thing with the other fraction. In this case, we multiplied 5 by 4 to get 20, so we need to multiply 2 by 4 to get an equal fraction.

2 x 4 = 8
2/5 = 8/20
Our fraction is 8/20.

Now that the fractions both have the same denominator, we can add them together easily.

5 + 8 = 13
5/20 + 8/20 = 13/20
Add the numerators together to get the final answer.

Therefore, 13/20 of the group's video games are either at Jerome or Mario's house.

Hope this helps!
5 0
3 years ago
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