If a < b, and c is positive, then ac

tell how you know that a system of two linear equations has more than one solution. how many solutions will such a system have?

riadik2000 [5.3K]

For a <span>system of two linear equations to have more than one solution, the lines representing the equations must coincide, i.e. must be the same or a multiple of each other. In that case, the system will have infinitely many solutions.
</span>

5 0

3 years ago

What is the length of the base of a triangle with height 18 m and area 99 square m?

Ivan

Use the equation for the area of a triangle:
$A = \frac{1}{2} hb$
where A = area, h = height of the triangle, and b = base of the triangle.

You know that h=18m and A=99 $m^{2}$ . Plug those values into the equation and solve for b, the base.
$A = \frac{1}{2} hb \\ 
99 = \frac{1}{2} (18)b \\ 
99=9b\\
b=11$

The base is 11m long.

8 0

3 years ago

Read 2 more answers

What is the equation of a line that passes through the point (8, 1) and is perpendicular to the line whose equation is y=−23x+5

elena-s [515]

Answer:

$y-1=\frac{1}{23} (x-8)$

Step-by-step explanation:

We can write the equation of a line in 3 different forms including slope intercept, point-slope, and standard depending on the information we have. We have a point (8,1) and a slope from the equation y=-23x+5. We will chose point-slope since we have a point and slope.

Point slope: $y-y_1=m(x-x_1)$

$m\neq -23$ in our new equation because it us perpendicular to it. This means we will need to change it into its negative reciprocal which is $m=\frac{1}{23}$ .

We will substitute $m=\frac{1}{23}$ and $x_1=8\\y_1=1$ .

$y-1=\frac{1}{23} (x-8)$ .

This is the equation of the line perpendicular to y=-23x+5 that crosses through (8,1).

5 0

3 years ago

For the curve with equation y = <img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" align=

bixtya [17]

Answer:

Below.

Step-by-step explanation:

a) First we find the derivative of the function, which gives us the gradient in terms of x:-

y = x^2 + 4x + 3

Gradient at x = dy/dx = 2x + 4.

When the curve cuts the x-axis, y = 0:

x^2 + 4x + 3 = 0

(x + 1)(x + 3) = 0

so x = -1 and -3 where the curve cuts the x axis.

The gradients at these point are therefore:

At x = -1 gradient = 2(-1) + 4 = 2

at x = -3 gradient = 2(-3) + 4 = -2.

b) (i) Where the tangent is parallel to x axis the gradient = 0

so 2x + 4 = 0 giving x = -2.

We now need to substitute x = -2 into the original function to find the y coordinate.

So the coordinates are (-2, (-2)^2 + 4(-2) +3))

= (-2, -1)

(ii) First find the slope of the line 6x + 3y = 7:

3y = -6x + 7

y = -2x + 7/3.

So the slope is -2.

Thus -2 = 2x + 4

2x = -6

x = -3

So the coordinates are (-3, (-3)^2 + 4(-3) + 3)

= (-3, 0).

7 0

3 years ago

X 1 у 3 4 6. 516<br>What is R

Dmitrij [34]

Answer:

the question is not correct. please rewrite the question.

7 0

3 years ago

If a < b, and c is positive, then ac < be and