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o-na [289]
2 years ago
10

The first triangle is dilated to form the second triangle. What id the scale factor?

Mathematics
1 answer:
Slav-nsk [51]2 years ago
7 0

Answer:

0.5

Step-by-step explanation:

10/5=0.5, or one half.

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How many solutions does this system of equations have?
JulsSmile [24]

Answer:

C. exactly one

Step-by-step explanation:

Hope this is helpful! Happy Halloween!

3 0
3 years ago
Aaron says that all
Softa [21]

Answer:

NO

Step-by-step explanation:

Look at the figure attached below, we know that the area of the cone is the sum of area of circle part and cone part of the figure.

to find the surface area of the cone, first measure the radius of the base and find the area of the circle part by  formula \pi r^2. Then measure the side (slant height) length of the cone part and find the area of the cone part by the formula \pi rl.

Now the surface area of the cone is circumference of the circle plus area of the cone part.

i.e. surface\ Area\ of\ the\ cone=\pi r^{2}+\pi rl

From the above discussion we concluded that surface area of the cone does not depend only on the circumference of the base but also we need side length of the cone part as well thus <em>all  cones with a base circumference of 8 inches  will </em><em>not </em><em>have the same surface area.</em>

<em />

5 0
3 years ago
Expand the following x(x-6), x(4x-1), 2x(5x+4), 3x(5x-y)
zlopas [31]

Answer:

x^2-6x, 4x^2-1x,10x^2+8x,15x^2-3xy

Just use the distributive property like you have before and you will get these answers, Hope this helped!

4 0
3 years ago
Read 2 more answers
Use lagrange multipliers to find the shortest distance, d, from the point (4, 0, −5 to the plane x y z = 1
Varvara68 [4.7K]
I assume there are some plus signs that aren't rendering for some reason, so that the plane should be x+y+z=1.

You're minimizing d(x,y,z)=\sqrt{(x-4)^2+y^2+(z+5)^2} subject to the constraint f(x,y,z)=x+y+z=1. Note that d(x,y,z) and d(x,y,z)^2 attain their extrema at the same values of x,y,z, so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.

The Lagrangian is

L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)

Take your partial derivatives and set them equal to 0:

\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}

Adding the first three equations together yields

2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43

and plugging this into the first three equations, you find a critical point at (x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right).

The squared distance is then d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43, which means the shortest distance must be \sqrt{\dfrac43}=\dfrac2{\sqrt3}.
7 0
3 years ago
Jackson is starting an exercise program. The first day he will spend 30 minutes on the treadmill. He will increase his time on t
yulyashka [42]

Answer:

40 minutes.

Step-by-step explanation:

  • time spent on the thread mill as a function of d(day) = t(d)

= 30 +2(d-1) minutes

[where d represents day number]

(when d is 1 the time spent should be 30

so substitute d=1 and check it will be 30 only.

on first day he will spend 30 minutes so, I added 30 .

and on every additional day( these additional days are excluding first day), he will increase the time by 2 minutes.so, I added 2(d-1) to initial 30 minutes)

  • so, T(6), the minutes he will spend on the treadmill on day 6=

=30+2(6-1)

=30+2(5)

=30+10

=40 minutes

3 0
3 years ago
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