<em><u>No</u></em><em><u>,</u></em><em><u> </u></em><em><u>because</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>convert</u></em><em><u> </u></em><em><u>meter</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>millimeter</u></em><em><u> </u></em><em><u>you'll</u></em><em><u> </u></em><em><u>have</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>multiply the length value by 1000</u></em><em><u> </u></em><em><u>which</u></em><em><u> </u></em><em><u>will</u></em><em><u> </u></em><em><u>give</u></em><em><u> </u></em><em><u>you</u></em><em><u> </u></em><em><u>3</u></em><em><u>5</u></em><em><u>0</u></em><em><u>0</u></em><em><u>.</u></em>
<em><u>3.5</u></em><em><u> </u></em><em><u>×</u></em><em><u> </u></em><em><u>1</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em>
<em><u>=</u></em><em><u> </u></em><em><u>3</u></em><em><u>5</u></em><em><u>0</u></em><em><u>0</u></em><em><u>.</u></em>
<em><u>have</u></em><em><u> </u></em><em><u>a</u></em><em><u> </u></em><em><u>great</u></em><em><u> </u></em><em><u>day</u></em><em><u>!</u></em>
Answer:
To Find => The value of x. in the equation below
=>

Step-by-step explanation:
We will cross Multiply the digits
=>20 \div 22 = x \div 20
=>We'll cross Multiply
=>

<em><u>Hence,</u></em><em><u> </u></em><em><u>1</u></em><em><u>0</u></em><em><u> </u></em><em><u>is </u></em><em><u>the</u></em><em><u> </u></em><em><u>required</u></em><em><u> </u></em><em><u>answer</u></em>
-4m-24=-8
(add 24 to both sides)
-4m=16
(divide by -4 on both sides)
m=-4
Answer:
both should
be equal to hope my work is correct or helps
Step-by-step explanation:
first one
-25<-3x+5 subtract 5 from both sides
-25-5=-30
5-5=0
then divide by -3
-30 divided by -3 =10
x=10
so it is equal to not less than
second one
-3x+5<-31
subtract 5 from both sides 5-5=0
-31 -5=-36
then divide both sides by -3
x=12
1
Answer:
Step-by-step explanation:
Given a general quadratic formula given as ax²bx+c = 0
To generate the general formula to solve the quadratic equation, we can use the completing the square method as shown;
Step 1:
Bringing c to the other side
ax²+bx = -c
Dividing through by coefficient of x² which is 'a' will give:
x²+(b/a)x = -c/a
- Completing the square at the left hand side of the equation by adding the square of half the coefficient x i.e (b/2a)² and adding it to both sides of the equation we have:
x²+(b/a)x+(b/2a)² = -c/a+(b/2a)²
(x+b/2a)² = -c/a+(b/2a)²
(x+b/2a)² = -c/a + b²/4a²
- Taking the square root of both sides
√(x+b/2a)² = ±√-c/a + b²/√4a²
x+b/2a = ±√(-4ac+b²)/√4a²
x+b/2a =±√b²-4ac/2a
- Taking b/2a to the other side
x = -b/2a±√√b²-4ac/2a
Taking the LCM:
x = {-b±√b²-4ac}/2a
This gives the vertex form with how it is used to Solve a quadratic equation.