In mathematics multiplying decimals is just like multiplying a whole number and ignoring the decimals., then just put the decimals in the answer – it is how many decimals does the two numbers combined.
First step:
Line the numbers on the right and do not a line the decimal points
26.2
× 16.43
Second Step:
Multiply the numbers starting on the right, just like multiplying a whole number and ignore the decimals.
26.2
<span> × 16.43</span>
786
1048
1572
<span>+ 262 </span>
430466
Third step: add the products
26.2
<span> × 16.43</span>
786
1048
1572
<span>+ 262 </span>
430.466 is the mathematician’s answer.
Using the two parallel line theorems we proved that ∠8 ≅ ∠4.
In the given question,
Given: f || g
Prove: ∠8 ≅ ∠4
We using given diagram in proving that ∠8 ≅ ∠4
Since f || g, by the Corresponding Angles Postulate which states that "When a transversal divides two parallel lines, the resulting angles are congruent." So
∠8≅∠6
Then by the Vertical Angles Theorem which states that "When two straight lines collide, two sets of linear pairs with identical angles are created."
∠6≅∠4
Then, by the Transitive Property of Congruence which states that "All shapes are congruent to one another if two shapes are congruent to the third shape."
∠8 ≅ ∠4
Hence, we proved that ∠8 ≅ ∠4.
To learn more about parallel line theorems link is here
brainly.com/question/27033529
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78.5= (3.14)r^2
25=r^2
5=r
So, the radius of a circle with an area is 78.5 cm^2 is 5 cm^2
Answer:
and 
Step-by-step explanation:
Given:
Equation 1:
Simplifying the above equation by dividing both sides by 100
Equation 2:
Simplifying the above equation by dividing both sides by 400.
Now the system of equation is:
(1a) 
(2a) 
Solving by elimination
Multiplying equation (2a) with 
(2b) 
Adding equations (1a) and (2b) in order to eliminate 
+
We get 
∴
Plugging in equation (2a).
Adding 
both sides.
∴
Answer:
x = 2
Step-by-step explanation:
These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.
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<h3>Squaring</h3>
The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.
The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.
x = -1: √(-1+2) +1 = √(3(-1)+3) ⇒ 1+1 = 0 . . . . not true
x = 2: √(2+2) +1 = √(3(2) +3) ⇒ 2 +1 = 3 . . . . true . . . x = 2 is the solution
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<h3>Substitution</h3>
Another way to solve this is using substitution for one of the radicals. We choose ...
Solutions to this equation are ...
u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution
The value of x is ...
x = u² -2 = 2² -2
x = 2 . . . . the solution to the equation
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<em>Additional comment</em>
Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.
