Y = 5x + 5
y - x = 5
You can use substitution to solve this system. Use y's expression (5x + 5) for y in the second equation to solve for x:
y - x = 5
5x + 5 + x = 5
6x + 5 = 5
6x = 0
x = 0
Substitute your value for x into one of the original equations to y:
y = 5x + 5
y = 5(0) + 5
y = 5
Finally, substitute both values into both original equations to check your work:
5 = 5(0) + 5 --> 5 = 5 <--True
5 - 0 = 5 --> 5 = 5 <--True
Answer:
x = 0
y = 5
Answer:
i am positive it is x+y+z
Step-by-step explanation:
i'm so sorry if this is wrong..hope this helped :)
The equation that represents the array (rectangles and area) multiplication model that sows two grey shaded columns of length one ninth each and three rows with dots of width one fourth each is option <em>a</em>
a) The equation with fractions two ninths times three fourths is equal to six thirty sixths
<h3>What is an array (area) multiplication model?</h3>
An array representation of a multiplication is a rectangular visual order of positioning of rows and columns that indicates the terms of a multiplication equation.
Please find attached the area model to multiply the fractions
The terms of the equation represented by the model are indicated by the two columns of length one ninth each shaded grey and the three rows of width one fourth each covered with dots, such that the equation can be presented as follows;
The equation that the model represents is therefore;
- The equation with fractions two ninths times three fourths is equal to six thirty sixths
Learn more about multiplication models here:
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Answer:
21
Step-by-step explanation:
f(x) =x^2 - 6x + 21
f(xl =0
Answer:
substitute 0 for x
f(0) = 0^2 - 6(0) + 21
0. = 0 - 0 + 21
0. = 21
Answer:
0 <=t<=21
Step-by-step explanation:
Projectile is Moving upwards on an interval of (0 to 21), if we plot Velocity vs Time and denote positive y-axis above 0 and negative y-axis below 0(for velocity), then from 0 to 21 t projectile is moving upwards and has positive velocity, when the projectile reaches the top of it's motion and returns back down to ground it's velocity is negative and is plotted below the y =0 (note that is for t > 21).
hence for the interval 0 <=t <=21 the instantaneous velocity is positive (Note, instantaneous velocity is also the derivative of the velocity or the slope ).
