What is the slope of a line parallel to the line whose<br> equation is y = 5x + 4?

Wesley office had already recycled 37 kilograms this year before starting a new recycling plan, and the new plan will have the o

borishaifa [10]

Answer:

37 kilograms alr recycled this year

Plan:

1 kilogram each week

After 13 weeks he would of recycled 13 kilograms

37+13=50

In total his office wesley's have recycled 50 kilograms of paper

Hope this helps

8 0

3 years ago

If y is the circumcenter is angle STU find each measure

snow_tiger [21]

Answer:

Measures are SV=9 units., SY=14 units, YW= $\sqrt75units$ , YW= $\sqrt27units$

Step-by-step explanation:

Given Y is the circumcenter of ΔSTU. we have to find the measures SV, SY, YW and YX.

As Circumcenter is equidistant from the vertices of triangle and also The circumcenter is the point at which the three perpendicular bisectors of the sides of the triangle meet.

Hence, VY, YW and YX are the perpendicular bisectors on the sides ST, TU and SU.

Given ST=18 units.

As VY is perpendicular bisector implies SV=9 units.

Also in triangle VTY

$YT^{2}=VY^{2}+VT^{2}$

⇒ $14^{2}=VY^{2}+9^{2}$

⇒ VY^{2}=115

As vertices of triangle are equidistant from the circumcenter

⇒ SY=YT=UY=14 units

Hence, SY is 14 units

In ΔUWY, $UY^{2}=YW^{2}+UW^{2}$

⇒ $14^{2}=YW^2+11^{2}$

⇒ $YW^2=196-121=75$ ⇒ YW= $\sqrt75units$

In ΔYXU, $UY^{2}=YX^{2}+XU^{2}$

⇒ $14^{2}=YX^2+13^{2}$

⇒ $YX^2=196-169=27$ ⇒ YW= $\sqrt27units$

Hence, measures are SV=9 units., SY=14 units, YW= $\sqrt75units$ , YW= $\sqrt27units$

4 0

3 years ago

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

eimsori [14]

Answer:

There are NO real roots for this equation. The only roots have imaginary parts and therefore cannot be represented on the real x-axis.

Step-by-step explanation:

We notice that the expression on the left of the equation is a quadratic with leading term $2x^2$ , which means that its graph is that of a parabola with branches going up.

Therefore, there can be three different situations:

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

2) if its vertex is below the x-axis, the parabola's branches are 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 have NO 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 $a=2$ and $b=-3$ , we get that the x-position of the vertex is:

$x_v=\frac{-b}{2a}\\x_v=\frac{-(-3)}{2(2)}\\x_v=\frac{3}{4}$

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = 3/4:

$y_v=f(\frac{3}{4})=2( \frac{3}{4})^2-3(\frac{3}{4})+4\\f(\frac{3}{4})=2( \frac{9}{16})-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{18}{8}+\frac{32}{8}\\f(\frac{3}{4})=\frac{23}{8}$

This is a positive value for y, therefore we are in the situation where there is NO x-axis crossing of the parabola's graph, and therefore no real roots.

We can though estimate a few more points of the parabola's graph in order to complete the graph as requested in the problem. For such we select a couple of x-values to the right of the vertex, and a couple to the right so we can draw the branches. For example: x = 1, and x = 2 to the right; and x = 0 and x = -1 to the left of the vertex:

$f(-1) = 2(-1)^2-3(-1)+4= 2+3+4=9\\f(0)=2(0)^2-3(0)+4=0+0+4=4\\f(1)=2(1)^2-3(-1)+4=2-3+4=3\\f(2)=2(2)^2-3(2)+4=8-6+4=6$