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weqwewe [10]
3 years ago
8

What is the slope of the line that passes through the points (8, 1) and (5, 7)?

Mathematics
1 answer:
ololo11 [35]3 years ago
7 0

Answer:

m= -2

Step-by-step explanation:

For future reference, you should try using Symbolab, it works really well and I use it ALL the time!

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There are 40 students in a class.
Anvisha [2.4K]

Step-by-step explanation:

Total no. of students = 40

Total no. of girls = 22

Total no. of boys = 40 - 22 = 18

No. of girls walking to school = 9

No. of girls who cycle to school = 22 - 9

No. of boys who cycle to school = 7

No. of students who take the bus = 10

(6 boys and 4 girls)

No. of boys who walk to school = 18 - (7 + 6) = 5

Total no. of students who walk to school = 9 + 5 = 14

5 0
3 years ago
How much volume is in a globe
Varvara68 [4.7K]

Answer:

904.32

Step-by-step explanation:

I did the math

3 0
3 years ago
Read 2 more answers
In the equation 4+c=18, the value of c will be less than 18 true or false?
Dominik [7]

Answer:

true

Step-by-step explanation:

because when we subtract 18 and 4, the answer is 12. Hence, the value is less than 18

5 0
3 years ago
Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)
leonid [27]
Let P=(x,y,z) be an arbitrary point on the surface. The distance between P and the given point (7,11,0) is given by the function

d(x,y,z)=\sqrt{(x-7)^2+(y-11)^2+z^2}

Note that f(x) and f(x)^2 attain their extrema, if they have any, at the same values of x. This allows us to consider the modified distance function,

d^*(x,y,z)=(x-7)^2+(y-11)^2+z^2

So now you're minimizing d^*(x,y,z) subject to the constraint z^2=xy. This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be

\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)

which has partial derivatives

\begin{cases}\dfrac{\mathrm d\mathcal L}{\mathrm dx}=2(x-7)-\lambda y\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dy}=2(y-11)-\lambda x\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dz}=2z+2\lambda z\\\\\dfrac{\mathrm d\mathcal L}{\mathrm d\lambda}=z^2-xy\end{cases}

Setting all four equation equal to 0, you find from the third equation that either z=0 or \lambda=-1. In the first case, you arrive at a possible critical point of (0,0,0). In the second, plugging \lambda=-1 into the first two equations gives

\begin{cases}2(x-7)+y=0\\2(y-11)+x=0\end{cases}\implies\begin{cases}2x+y=14\\x+2y=22\end{cases}\implies x=2,y=10

and plugging these into the last equation gives

z^2=20\implies z=\pm\sqrt{20}=\pm2\sqrt5

So you have three potential points to check: (0,0,0), (2,10,2\sqrt5), and (2,10,-2\sqrt5). Evaluating either distance function (I use d^*), you find that

d^*(0,0,0)=170
d^*(2,10,2\sqrt5)=46
d^*(2,10,-2\sqrt5)=46

So the two points on the surface z^2=xy closest to the point (7,11,0) are (2,10,\pm2\sqrt5).
5 0
4 years ago
In triangle abc, m of acb = 90, cd is perpendicular to ab , m of acd is 60. and bd is 5 cm. find ad
weeeeeb [17]

Let us draw a picture to make the things more clear.

Attached is the image.

We have been given that

\angle acd = 60 ^{\circ}

Therefore, we have

\angle dcb =90- 60= 30 ^{\circ}

Now, in triangle bcd, we have

\tan30 = \frac{5}{cd}\\
\\
\frac{1}{\sqrt 3}=\frac{5}{cd}\\
\\
cd=5\sqrt 3

Now, in triangle acb, we have

tan 60 = \frac{ad}{5\sqrt3} \\
\\
\sqrt 3=  \frac{ad}{5\sqrt3}\\
\\
ad= 5\sqrt3 \times \sqrt 3\\
\\
ad= 5\times 3\\
\\
ad=15

Thus, ad is 15 cm.


4 0
3 years ago
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