Let 'x' be the problems worth 2 points.
Let 'y' be the problems worth 3 points.
Since, there are 38 total problems.
So, 
(equation 1)
x = 38-y
Since, a perfect score is 100 points.
So, 
(equation 2)
Substituting the value of 'x', we get
y = 24
x+y = 38
x = 38-24 = 14
So, 14 problems are worth 2 points and 24 problems are worth 3 points.
Answer:
Step-by-step explanation:
an equation is : Use the point-slope formula.
y - y_1 = m(x - x_1) ; m : the slope when : x_1 = -3 and y_1 = 7
- 2 ×m = - 1 because this line is perpendicular to the line y= -2x-5 when the slope is -2
so : m= 1/2
an equation is : y - 7 =(1/2)(x+3)
Please provide an image or other when you ask a question, as you’re not gonna get a clear answer
Ayanna is 17 and Brittany is 14.
To find this, let's write two equations using the given information.
a = b + 3
3b + 2a = 76
Now, just substitute the first equation into the second and solve.
3b + 2(b + 3) = 76
3b + 2b + 6 = 76
5b = 70
b = 14
If Brittany is 14, then Ayanna is 3 years older at 17.
First one:
you can add -10m and -13m but you can't add -10m and 2m^4 becuase the powers aren't the same so
when adding the like terms
look at the:
powers, (x^3 adds with x^3)
placehloder letter (x adds with x and y adds with y and so on)
-10m+2m^4-13m-20m^4
powers: m^1 and M^4
placeholders: all m
add
-10m-13m+2m^4-20m^4
-23m-18m^4
second one:
when multiplying exponents, you add with like
so if you multipliy
x^2yz^3 times x^4y^2z^2 thne you would get x^6y^3z^5
when multiply with coeficients
2x^2yz^3 times 4x^4y^2z^2=8x^6y^3z^5
so using associative property a(bc)=(ab)c
2/3 times p^4 times y^3 times y^4 times s^5 times 6 times p^2 times s^3
group like terms
(2/3 times 6) times (p^4 times p^2) times (y^3 times y^4) times (s^5 times s^3)
(4) times (p^6) times (y^7) times (s^8)
4p^6y^7s^8
