Please help with this()

Answers

Answer 1

Answer:

Answer:

396

Step-by-step explanation:

Related Questions

20. The temperature of a pot of water is 62*F. The temperature increasesby 20*F per minute when being heated. Write an equation to representthe linear relationship. *
HELP!!!!! PLEASE!!!!!
D (t)=16t□2+96t+112 where t represents time in seconds
A cold front changes the temperature by -3 F each day. If the temperature started at 0 F, what will the temperature be after 5 days?
During his tennis career in singles play, John won 3 fewer tournament A titles than tournament B titles, and 2 more tournament C titles than tournament B titles. If he won 17 of these titles total, how many times did he win each one? How many A titles How many B titles How many C titles

Santa can see up to 25 children per hour what is the maximum number of children he can see in 6 hours?

Answers

The required Santa cane sees up to 150 children in 6 hours.

What is arithmetic?

In mathematics, it deals with numbers of operations according to the statements. There are four major arithmetic operators, addition, subtraction, multiplication and division,

What is simplification?

The process in mathematics to operate and interpret the function to make the function or expression simple or more understandable is called simplifying and the process is called simplification.

Here,
Santa can see up to 25 children per hour,
Now,
The number of children that Santa can see in 6 hours is given as,
= 25 × 6
= 150 children

Thus, the required Santa cane sees up to 150 children in 6 hours.

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He can see up to 150 because, 25x6 is 150 children.

The height , H , in feet of an object above the ground is given by h(t)=-16t^2+64t+190 , where t is the time in seconds . What is the height of the object after 2 seconds

Answers

After 2 seconds.
h(2) = -16(2)^2 + 64(2) + 190 = -16(4) + 128 + 190 = -64 + 128 + 190 = 254 feet

Final answer:

The height of the object after 2 seconds is 254 feet.

Explanation:

The height of an object after 2 seconds can be found by substituting t = 2 into the equation for height:

h(2) = -16(2)² + 64(2) + 190

h(2) = -64 + 128 + 190

h(2) = 254

Therefore, the height of the object after 2 seconds is 254 feet.

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7a-3=3-2a
I don't know how to solve this question

Answers

7a-3=3-2a
9a-3=3 (add 2a to both sides)
9a=6 (add 3 to both sides)
a=6/9 (divide by 9 to both sides)

get all variables (in this case, "a") on one side by adding "2a" to both. this cancels the "-2a". then, combine like terms. also move "-3" to the other side, by adding "3". finally, divide the "6" and "a" by "9" to get variable alone. therefore, "a = 2/3".
7a-3=3-2a
+2a +2a
9a-3=3
+3 +3
9a=6
a=2/3

Equations with the variable on both sides . solve 2+ 6x = 1-x

Answers

so 2+6x=1-x
we can do anything as long as we do it to both sides
2+6x=1-x
try to get the unknown (x) on one side by itself
add x to both sides to get rid of the x on the right
2+x+6x=1+x-x
2+7x=1
subtract 2 from both sdies
7x=-1
divide by 7
x=-1/7

2 + 6x = 1 - x
+ 6x + x = 1 - 2
7x = - 1
x = - 1
7

Solution: -1/7

A square has an area of 49 square meters. What is the perimeter of the square? 1) 28 meters 2) 30 meters 3) 32 meters 4) 36 meters

Answers

Answer:

1)28

Step-by-step explanation:

The area of square=49m^2

or,L^2=49

i.e. L=7

also perimeter =4L

or, =4×7

or, perimeter=28m

Which relationships hold true for the sum of the magnitudes of vectors u and v, which are perpendicular? a. ||u+v||=||u||+||v|| b. ||u+v||=sq rt(||u||^2+||v||^2) c. ||u+v||

Answers

The perpendicular are B) ||u+v|| = sqrt ||u||^2 + ||v||^2 and D) ||u+v|| less than ||u||+||v||

We have given that,

The relationships hold true for the sum of the magnitudes of vectors u and v,

We have determined which are perpendicular.

What is the magnitude of the vector?

The magnitude of a vector formula is used to calculate the length for a given vector (say v) and is denoted as |v|. So basically, this quantity is the length between the initial point and endpoint of the vector.

B) ||u+v|| = sqrt ||u||^2 + ||v||^2

AND

D) ||u+v|| less than ||u||+||v||

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it is

B) ||u+v|| = sqrt ||u||2 + ||v||2

AND

D) ||u+v|| less than ||u||+||v||