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3 years ago

Which quantity is not equal to 75(32+88)

Mathematics

1 answer:

Anna35 [415]3 years ago

3 0

Without the choices given it would be any number
EXCEPT 9000
75(32+88)=
75(120)= 9000

HOPE THIS HELPS

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A line passes through the point (8,7) and has a slope of negative 3/4 Write an equation in slope-intercept form for this line.

Anton [14]

Answer: Y= -3/4x+13

Step-by-step explanation:

A line passes through the point (8,7) and has a slope of negative 3/4 Write an equation in slope-intercept form for this line.

This is what i get

Slope of a line Y= mx+b

Y= 7

X= 8

M= -3/4

7= -3/4(8)+b

7= -6+b

7+6=b

13=b

Y= -3/4x+13

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3 years ago

____ [38]

Answer:

y = 5x+3

Step-by-step explanation:

a general line is represented by the equation:

y = mx + n

a line with m=5, that goes through (-1,-2), should satisfy the equation:

-2 = 5×(-1) + n

this is because any point on the line, satisfies it's equation.

from this we get n = 3, hence our answer

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2 years ago

What is the area of 5m, 17m, 8m, and 5m, the shape is a kite

Andrei [34K]

Answer:

It can't be a kite. A kite has 2 pairs of equal sides.

Step-by-step explanation:

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3 years ago

Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat

zavuch27 [327]

Answer:

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.