Answer:
the length of base triangle is 42cm
The place value of 9 in 0.496 is hundredths place.
1.) 4(x+3)
Find the GCF, Greatest Common Factor, of 4x and 12.
4x=2*2*x
12=3*2*2
The greatest common factor is 4. Put this outside of the parentheses. (You would multiply the 2*2)
Then, put the rest of the factors as a sum. (Only the factors on the same line.)
Solution: 4(x+3)
To check, distribute to see if it works.
4x+12
2.) 2(4r+7)
Find the GCF of 8r and 14
8r=2*2*2*r
14= -1*7*2
The greatest common factor is 2. (There is only 1 two, so you would not multiply them.)
Then, put the rest of the factors as a sum. (Only the factors on the same line.)
Multiply the 2*2*r as one addend and the -1*7 as the other.
Solution: 2(4r-7)
To check, distribute to see if it works.
8r-14
Do you get it now?
3.) 5(x+7)
4.) 7(2x+1)
5.) Cannot be factored.
32x-15
Find the GCF of 32x and -15
32x: 2*2*2*2*2*x
-15: -1*5*3
Because there are no similar factors other than 1, it cannot be factored.
6.) 8(4x+3)
7.) 3(2x-3)
8.) 24(1x+2)
9.) 9(-2x+8)
10.) Cannot be factored
11.) 8(1x+3)
12.) 50(1x+5)
Answer:
29. See table below
30. See attached graph
31. The slope is m= 0.10
The slope represent the cost for every additional call minute.
Step-by-step explanation:
The cost is $0.5 first minute and $0.10 for any additional minutes
If c is the total cost of a call that last t minutes then;
c= 0.10t + 0.5-----where t is the time the call lasted
29. Use the equation above to create the table as;
t {x} c{y}
1 0.6
2 0.7
3 0.8
4 0.9
5 1.0
6 1.1
The graph of this plot is as attached , where the coordinates are
{1,0.6} , {2,0.7} ,{3,0.8} ,{4,0.9} ,{5,1.0}, {6,1.1}
The slope can be found using the formula;
m=Δy/Δx
m= 1.1 - 0.6 / 6-1
m= 0.5 / 5 = 0.10
The slope represent the cost for every additional call minute.
We know that
Limacons are in the form
r=a(+-)b*sin (theta) or r=a(+-)b*cos (theta)
in this problem
r=a(+-)b*sin (theta)
so
is symmetric to y axis
a=1/2
b=1/2
a/b=1---------> Cardioid (hearth shaped)
see the attached figure
the answer is the option
<span>
A. Cardioid </span>