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Amanda [17]
3 years ago
11

Please help me with the question :)

Mathematics
1 answer:
poizon [28]3 years ago
3 0

SV middle line of trapezoid, so

\frac{40 + x}{2}  = 29 \\ 40 + x = 58 \\ x = 18

Answer: QR=18

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What is the point of intersection when the system of equations below is graphed on the coordinate plane?
lora16 [44]

Answer:

<h2>not exist</h2>

Step-by-step explanation:

The coordinates of the intersection of the line are the solution of the system of equations.

\underline{+\left\{\begin{array}{ccc}x-y=1\\y-x=1\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad\qquad0=2\qquad\bold{FALSE}

The system of equations has no solution. Therefore, the lines are parallel (the intersection does not exist).

5 0
3 years ago
What numbers would you multiply -3/4 by to equal 1?
Sonja [21]
You would multiply by -4/3. That's cuz to get to 1, you have to cancel out the negative, with another negative. You then have to multiply by the reciprocal. 

Hope this helps! 
7 0
3 years ago
What is the middle term in the simplified product? <br> (x + 3)(x – 4)
Eduardwww [97]
The middle term is -x
4 0
3 years ago
Read 2 more answers
A Ferris wheel has a radius of 10 m, and the bottom of the wheel passes 1 m above the ground. If the Ferris wheel makes one comp
Volgvan

Answer:1+10(1-\cos (\frac{\pi t}{9}))

Step-by-step explanation:

Given

radius of wheel r=10 m

Time period of Wheel T=18 s

and T\cdot \omega =2\pi , where \omega =angular velocity of wheel

\omega =\frac{2\pi }{18}

Let at any angle \thetawith vertical position of a point is given by

x=r\sin \theta

y=y_0+r(1-\cos \theta )

and \theta =\omega \times t

for velocity differentiate x and y to get

v_x=r\cos \theta =r\cos (\omega t)

v_y=0+r(\sin \theta )=r\sin (\omeag t)

Height at any time t is given by

h=1+10(1-\cos \theta )=1+10(1-\cos (\frac{\pi t}{9}))

7 0
3 years ago
Please help I will mark brainliest if right
o-na [289]

Answer:

3

x

+

3

=

24

x

2

−

10

x

≤

−

9

−

2.5

x

+

5

=

3

Step-by-step explanation:

3 0
3 years ago
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