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Kay [80]
3 years ago
14

Write an equivalent expression for each of the following please: a. x+x+x+2x+x=. b.(x+2)+(x+2). c. y+y+y+y+10+y+1

Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
5 0

Answer:

A 6x

B 2x+4

C 5y + 11

Step-by-step explanation:

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Y = (2/3)x - 1<br> y = -x + 4
Vinil7 [7]
Are these two separate questions?
7 0
3 years ago
Geometry Select the correct answer from each drop-down menu. Given that , ° and °.
Alexxx [7]

\measuredangle BAC = 29°

Answer:

\measuredangle BOC =58°

Step-by-step explanation:

\measuredangle BDC = 29°...(Given) \\\because \measuredangle BAC = \measuredangle BDC\\(\angle 's \: subtended \: by\: same\: arc) \\\huge \purple {\boxed {\therefore \measuredangle BAC = 29°}} \\\\

Since, angle subtended on the center of the circle is twice the angle subtended on the circumference of the circle.

\therefore \measuredangle BOC =2\times \measuredangle BDC\\\therefore \measuredangle BOC =2\times 29°\\\huge \red {\boxed {\therefore \measuredangle BOC =58°}} \\

8 0
3 years ago
A ball is thrown into the air. The height in feet of the ball can be modeled by the equation h = -16t2 + 20t + 6 where t is the
makkiz [27]

f(t)=h=-16t^2+20t+6&#10;&#10;

a=-16, b=20, c =6

Δ=b^2-4ac=20^2-4*(-16)*6=400+384=784

\sqrt{Δ}=\sqrt{784}=28

max:

p=-\frac{b}{2a}=-\frac{20}{-32}=\frac{20}{32}=\frac{5}{8}

max heigh: h=-16(\frac{5}{8})^2+20*\frac{5}{8}+6=-16\frac{25}{64}+\frac{100}{8}+6

h=-\frac{25}{4}+\frac{50}{4}+6=\frac{-25+50+24}{4}=\frac{49}{4}

h=\frac{49}{4} at t=\frac{5}{8}s


will fall down at t:

t=\frac{-20-28}{-32}=\frac{-48}{-32}=\frac{3}{2}=1.5

t=1.5s

8 0
3 years ago
To join a Yoga club there is a $100 annual fee and a $5 fee for each class you attend.
Leona [35]

Answer:

210

Step-by-step explanation: so 5 times 12 120 plus 100 is 210

5 0
3 years ago
Which of the following is equivalent to –2i(6 – 7i)? (0 – 2i)(6 – 7i) (0 + 2i)(6 – 7i) (–2 + i)(6 – 7i) (6 – 2i) – (7i – 2i)
Mariulka [41]

Answer:

The expression which is equivalent to –2i(6 – 7i) is; (0 – 2i)(6 – 7i)

4 0
2 years ago
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