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3 years ago

20 pts + BRAINLIEST

Mathematics

1 answer:

exis [7]3 years ago

7 0

N 4
Part A
a) 7a+2a+3b-------> (7a+2a)+3b-----> 9a+3b
b) 8p+2p-7q-------> (8p+2p)-7q------> 10p-7q
c) 9a²+2a²+5a-----> (9a²+2a²)+5a---> 11a²+5a
d) 5ab+2ab-7a-----> (5ab+2ab)-7a---> 7ab-7a
e) 7+2a-a------> 7+(2a-a)-----> 7+a
f) 2a+3b+5a+2b---> (2a+5a)+(3b+2b)-----> 7a+5b
g) 6p+2q-4p+4q---> (6p-4p)+(2q+4q)----> 2p+6q
h) 9m²+7m²+5m-3m-----> (9m²+7m²)+(5m-3m)------> 16m²+2m
i) 7a²+4a-2a²+7----> (7a²-2a²)+4a+7------> 5a²+4a+7
j) 10p-3q+p-7p-----> (10p+p-7p)-3q------> 4p-3q
k) 3x²+2x-2x²+3x+x²---> (3x²-2x²+x²)+(2x+3x)-------> 2x²+5x

N 5
a) If Julie buys 4 boxes and Michelle buys 7 boxes
in total
(4+7)*n-----> 7*n
the answer Part 5 a) is 7*n

b) (5kg+3kg)*b---------> 8*b
the answer Part 5 b) is 8*b

c) (3+4+6)*x-------> 13*x
the answer Part 5 c) is 13*x

N 6
a) L+B+L+B------> (L+L)+(B+B)-----> 2*L+2*B-----> 2*(L+B)

b)
1) L+L+L-----> 3*L
2) 5+5+5+5-----> 5*4
3) x+x+y-----> 2*x+y
4) a+b+c+c----> a+b+2*c

N 7
a) s+s+s+s+s+s+s+s------> 8*s
b) t+t+t+t-----> 4*t
c) t+t+t+t+t+t----> 6*t
d) t+t+t+s+s----> 3*t+2*s
e) t+t+t+s+s+s----> 3*t+3*s
f) 12*t+6*s
g) 10 squares+12 triangles-----> 10*s+12*t
h) n squares+m triangles-----> n*s+m*t

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The equation contains exact roots at x = -4 and x = -1.

See attached image for the graph.

Step-by-step explanation:

We start by noticing that the expression on the left of the equal sign is a quadratic with leading term $x^2$ , which means that its graph shows branches going up. Therefore:

1) if its vertex is ON the x axis, there would be one solution (root) to the equation.

2) if its vertex is below the x-axis, it is forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will not have real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently:

We recall that the x-position of the vertex for a quadratic function of the form $f(x)=ax^2+bx+c$ is given by the expression: $x_v=\frac{-b}{2a}$

Since in our case $a=1$ and $b=5$ , we get that the x-position of the vertex is: $x_v=\frac{-b}{2a} \\x_v=\frac{-5}{2(1)}\\x_v=-\frac{5}{2}$

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = -5/2:

$y_v=f(-\frac{5}{2})\\y_v=(-\frac{5}{2} )^2+5(-\frac{5}{2} )+4\\y_v=\frac{25}{4} -\frac{25}{2} +4\\\\y_v=\frac{25}{4} -\frac{50}{4}+\frac{16}{4} \\y_v=-\frac{9}{4}$

This is a negative value, which points us to the case in which there must be two real solutions to the equation (two x-axis crossings of the parabola's branches).

We can now continue plotting different parabola's points, by selecting x-values to the right and to the left of the $x_v=-\frac{5}{2}$ . Like for example x = -2 and x = -1 (moving towards the right) , and x = -3 and x = -4 (moving towards the left.

When evaluating the function at these points, we notice that two of them render zero (which indicates they are the actual roots of the equation):

$f(-1) = (-1)^2+5(-1)+4= 1-5+4 = 0\\f(-4)=(-4)^2+5(-4)_4=16-20+4=0$

The actual graph we can complete with this info is shown in the image attached, where the actual roots (x-axis crossings) are pictured in red.

Then, the two roots are: x = -1 and x = -4.

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