Scientific notation means to write the answer in standard form. standard form helps you to express numbers that are too big or too small to be conveniently written in decimal form.
so for this case, it will be 4.571×10^-4.
when the number is decimal, you move the dot to the right and it will always be multiplying 10 to the power of a negative power.
ANSWER
EXPLANATION
The given matrix is a 4 by 4 matrix.
The entry in the intersection of the first third row and the fourth column is denoted
This entry is 8.
Check the attachment for how we got the 8.
Therefore
Answer:
There are no extraneous solutions
Reasoning:
An extraneous solution is a solution that isn't valid, it might be imaginary like the square root of a negative number.
first we want to isolate z:
1+sqrt(z)=sqrt(z+5)
^2 all ^2 all
(1+sqrt(z))(1+sqrt(z))=z+5
expand
1+2sqrt(z)+z=z+5
-1 -z -z -1
2sqrt(z)=4
/2 /2
sqrt(z)=2
^2 all ^2 all
z=4
Since there is one solution and it is a real number, there are no extraneous solutions.
Answer:
Problem 1:
Problem 2:
Problem 3:
The radius is cm.
Problem 4:
The width is 15 cm.
Step-by-step explanation:
Problem 1:
We want to solve for .
Multiply both sides by 3:
Rearrange the multiplication using commutative property:
We want to get by itself so divide both sides by what is being multiplied by which is .
Problem 2:
We want to solve for in .
Multiply both sides by 3:
We want by itself so divide both sides by what is being multiply by; that is divide both sides by .
Problem 3:
The circumference formula for a circle is . We are asked to solve for the radius when the circumference is cm.
Divide both sides by what r is being multiply by; that is divide both sides by :
Reduce fraction:
The radius is cm.
Problem 4:
The perimeter of a rectangle is where is the width and is the length.
We are asked to find w, the width, for when L, the length, is 5, and the perimeter is 40.
So we have this equation to solve for w:
Simplify the 2(5) part:
Subtract both sides by 10:
Divide both sides by 2:
Simplify the fraction:
The width is 15 cm.
It should be the third root of 24 or 2 on the outside of the third root of 3