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GarryVolchara [31]
3 years ago
6

Uh do yall know the coordinates im supposed to put for the last one

Mathematics
2 answers:
4vir4ik [10]3 years ago
5 0

4,-5

look at the differences between the two on the left and use them for the right. ( 3 to the left and 5 up)

(dk this topic just what I think)

tatiyna3 years ago
3 0
I agree with the user that’s above me
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FIND THE SURFACE AREA HELP ASAP GET BRAINLISTS
3241004551 [841]

Answer:

\large\boxed{a.\ 7200\pi\ mm^2}

Step-by-step explanation:

The formula of a surface area of a cylinder:

S.A.=2\pi r(r+H)

r - radius

H - height

We have r = 40mm and H = 50mm. Substitute:

S.A.=2\pi(40)(40+50)=80\pi(90)=7200\pi\ mm^2

4 0
4 years ago
15. If y varies directly as x and y = 540 when x = 10, find x when y = 1080
Sholpan [36]

\qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad \stackrel{\textit{constant of variation}}{y=\stackrel{\downarrow }{k}x~\hfill } \\\\ \textit{\underline{x} varies directly with }\underline{z^5}\qquad \qquad \stackrel{\textit{constant of variation}}{x=\stackrel{\downarrow }{k}z^5~\hfill } \\\\[-0.35em] \rule{34em}{0.25pt}

\stackrel{\textit{"y" varies directly as "x"}}{y = kx}\qquad \textit{we also know that} \begin{cases} y = 540\\ x = 10 \end{cases}\implies 540=k(10) \\\\\\ \cfrac{540}{10}=k\implies 54=k~\hfill \boxed{y=54x} \\\\\\ \textit{when y = 1080, what is "x"?}\qquad 1080=540x\implies \cfrac{1080}{540}=x\implies 2=x

6 0
2 years ago
What is the hypothenuse of a right triangle with bases 13cm and 5cm?
NikAS [45]
To find the hypothenuse use Pythagorean theorem: 13²+5²=h²,
h=√13²+5²=13.93cm = hypothenuse.
5 0
3 years ago
A house is built on coordinates (2,9). There is a water pipeline that runs along the line y = 2 3 x – 1. What is the approximate
Basile [38]

Answer:

7.21 units

Step-by-step explanation:

The shortest length needed to connect the house to the existing pipeline is the perpendicular distance between the location of the house (2,9) and the line y = 2/3x - 1.

The perpendicular distance (d) between a point (x_1,y_1) and the line Ax + By + C  = 0 is given as:

d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2} }

The line is given by y = 2/3 x - 1; 2/3x - y - 1 = 0. The point = (2, 9)

hence A = 2/3, B = -1, C = -1, x_1=2,y_1=9\\. Therefore:

d=\frac{|\frac{2}{3}(2)+(-1)(9)+(-1) |}{\sqrt{\frac{2}{3}^2+(-1)^2 } } \\\\d=7.21

4 0
3 years ago
A class of 4th graders sold tickets to their school play. Each of the 23 students in the class sold 4 tickets. The cost of each
zaharov [31]

Answer:

The made $644

Step-by-step explanation:

23*4= 92 (how many ticks sold all together)

92*$7= $644 (the number of tickets multiplied by the price)

8 0
3 years ago
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