24 POINTS!!!!!!!!!!!!!!!!!! Factor the expression. 4d + 12

A pole-vaulter uses a 15-foot-long pile. She grips the pole so that the segment below her left hand is twice the length of the s

Shalnov [3]

Let x + 1.5 be length of segment above left hand
Let 2x be length of segment below left hand

15 = x + 1.5 + 2x
15 = 3x + 1.5
15 - 1.5 = 3x
13.5 = 3x
13.5/3= 3x/3
4.5 = x
x = 4.5
This is the length above the left hand.

2x must be then 4.5*2 = 9 foot
This is the length below the left hand.

Checking
4.5 + 9 + 1.5 = 15 foot

Adding length below left hand & length of left hand to right hand:
9 + 1.5 = 10.5
The right hand is hence 10.5 foot far up the pole.

6 0

3 years ago

If a bicyclist traveled at 8:am to 8:18am a distance of 4.5 miles and from 8:18am to 8:45am a distance of 7.5 miles .What is the

Vinil7 [7]

8 am at 0

8:18 at 4.5

8:48 at 7.5 miles ???

18 min = .3 hour

48 min = .8 hour

in first .3 hours speed = 4.5/.3 = 15 mph

in next .5 hours = speed = 3/.5 = 6 mph

average speed for the whole trip = 7.5 miles/.8 hours = 9.375 mph

3 0

3 years ago

If you were to write $190 as either $19 or $1,900, this mistake would be called a(n)

Verizon [17]

It would be called a mess up on ur paper

3 0

3 years ago

A flat circular plate has the shape of the region x squared plus y squared less than or equals 1.The plate, including the bound

rjkz [21]

Answer:

We have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . and the hottest value is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

Step-by-step explanation:

We need to take the derivative with respect of x and y, and equal to zero to find the local minimums.

The temperature equation is:

$T(x,y)=x^{2}+2y^{2}-\frac{3}{2}x$

Let's take the partials derivatives.

$T_{x}(x,y)=2x-\frac{3}{2}=0$

$T_{y}(x,y)=4y=0$

So, we can find the critical point (x,y) of T(x,y).

$2x-\frac{3}{2}=0$

$x=\frac{3}{4}$

$4y=0$

$y=0$

The critical point is (3/4,0) so the temperature at this point is: $T(\frac{3}{4},0)=(\frac{3}{4})^{2}+2(0)^{2}-(\frac{3}{2})(\frac{3}{4})$

$T(\frac{3}{4},0)=-\frac{9}{16}$

Now, we need to evaluate the boundary condition.

$x^{2}+y^{2}=1$

We can solve this equation for y and evaluate this value in the temperature.

$y=\pm \sqrt{1-x^{2}}$

$T(x,\sqrt{1-x^{2}})=x^{2}+2(1-x^{2})-\frac{3}{2}x$

$T(x,\sqrt{1-x^{2}})=-x^{2}-\frac{3}{2}x+2$

Now, let's find the critical point again, as we did above.

$T_{x}(x,\sqrt{1-x^{2}})=-2x-\frac{3}{2}=0$

$x=-\frac{3}{4}$

Evaluating T(x,y) at this point, we have:

$T(-(3/4),\sqrt{1-(-3/4)^{2}})=-(-\frac{3}{4})^{2}-\frac{3}{2}(-\frac{3}{4})+2$

$T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$

Now, we can see that at point (3/4,0) we have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . On the other hand, at the point $-(3/4),\frac{\sqrt{7}}{4})$ we have the hottest value of temperature, it is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

I hope it helps you!

4 0

2 years ago

On Monday Harold picked up four donuts and two large coffees for the office staff. He paid 4.08. On Tuesday Melinda picked up tw

Mandarinka [93]

One way to look at this is to make 2 equations and treat it as a set of equations. We will make the cost of a donut equal to x, and the cost of a coffee equal to y.

Harold's order:
4x + 2y = 4.08

Melinda's order:
2x + 3y = 3.92

To combine these equations, we need to cancel out either x or y:
4x + 2y = 4.08
2x + 3y = 3.92

4x + 2y = 4.08
4x + 6y = 7.84

4x + 2y = 4.08
-4x - 6y = -7.84

We can now combine these equations:
4x + 2y = 4.08
-4x - 6y = -7.84

-4y = -3.76

y = 0.94

Therefore the cost of a large coffee is $0.94. We can plug this into either original equation to find the cost of a donut.
4x + 2y = 4.08
4x + 2(0.94) = 4.08
4x + 1.88 = 4.08
4x = 2.20
x = 0.55

The cost of a donut is $0.55, and the cost of a large coffee is $0.94.

8 0

3 years ago