Make doubles.add. 5+6 __+_+_ so,5+6=11

2 answers:

5 0

If you want 2 numbers, you have: 1+10, 2+9, 3+8, and so on. If you need 3 numbers you have 1+1+9, 1+2+8, 2+2+7, and so on.

mario62 [17]3 years ago

3 0

The doubles plus one are the facts in which one addend is larger than the other by one. Examples 3+4=7, 6+7=13, 5+6=11, 3+2=5. Learning your doubles plus one should involve knowing the doubles and mentally adding the additional one. ubles are adding the same number together, as with 5+5=10.
With doubles +1, the numbers must be in beside each other in counting order going up or down (ex: 4, 5, 6 and 9, 8, 7). I teach my first graders to say the SMALLEST number, double it, then add one. So, 9+8 would be 8+8=16, +1=17. You must know your doubles facts before this strategy is useful.

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Y^2 + 18y - 41 = 0<br> Y= <br> Y=

lana66690 [7]

Answer:

y= -9+√122, or -9 - √122

Step-by-step explanation:

You have to solve the equation for y to find each variable of the quadratic and applying the quadratic formula.

after solved, I got the answer as decimal,

y = 2.0453610.....

i simpler form : y = 2.05

Also, an answer can be y= -9+√122, or -9 - √122

6 0

3 years ago

Example 2:

Montano1993 [528]

Taking this example into account, we can see that setting the first value equal to 1, we obtain that $F(x)=0.5x+1=4$ and $x=6$ . Using this information, we find that $F(x+1)=0.5(x+1)+1=0.5(6+1)+1=4.5$ . It shows that when x is positive, the succussive terms are increasing.

Referring to that finding, if we set initial value less than zero, which means that we are solving 0.5x+1<0 and taking a number in the interval of the solution, which means x ∈ (- ∞, -20). Setting x=-19, we find that $F(x)=0.5x+1=-19$ and x=-40. In the next iteration, $F(x+1)=0.5(x+1)+1=0.5(1-40)+1=-18.5$ . In the next iteration, $F(x+2)=0.5(x+2)+1=0.5(2-40)+1=-18$ . By this way, we find that even if the initial value is less than zero, value of the successive iterations is increasing.

Using the function $g(x)=-x+2$ and taking the initial value equal to 4, we find that $g(x)=-x+2=4$ and x=-2. In the next iteration, $g(x+1)=-(x+1)+2=-(-2+1)+2=3$ . If we continue the iterations we'll see that they are decreasing.
Setting the initial value equal to 2, we find that $g(x)=-x+2=2$ and x=0. The next iteration is $g(x+1)=-(x+1)+2=1$ . In this case, the interations are also decreasing.
If we set the initial value equal to 1, we find that $g(x)=-x+2=1$ and x=1. In the next iteration, $g(x+1)=-(x+1)+2=0$ and the iterations are decreasing.