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Marrrta [24]
3 years ago
14

Q and T are the midpoints of the legs, RV and SU , of trapezoid RSUV. If RS=3y+12, QT=y+25, and UV=2y+32, what is the value of y

?

Mathematics
1 answer:
Anna71 [15]3 years ago
8 0
100 bc I did the work
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- (A) The following data was received from the Student Affairs Unit of the University
ladessa [460]

Answer: i) 30 students for chemistry

ii) 21 students for only one subject

iii) 61 students

Step-by-step explanation: from the question we got 30 students registered for chemistry. number students only for biology = (27-18) = 9

number of students only for chemistry = (30-18) = 12

number of student registered for only one subject = (12+9)= 21

number of students registered for neither two subjects = {100-(9+12+18)} = 61

5 0
3 years ago
Find the value of x.<br> B<br> 10-X<br> 3<br> D<br> х<br> 2.<br> A<br> С<br> x = [?]
Nuetrik [128]

Answer:

3/(10-x) = 5/(10-x+x)

3/(10-x) = 1/2

6 = 10-x

x = 4

Hope this helps!

7 0
3 years ago
PLS Help!!!..........
oksano4ka [1.4K]

Answer:

A) 0.50x +300 < 650, where x < 700

(Please mark brainliest. :3)

5 0
2 years ago
Read 2 more answers
4(x+3)-4=8(1/2x+1) i need to find the solution
oee [108]
You can solve this by algebraically as well as graphically. When I tried both ways I got:  
solution is x= -2<span />
5 0
3 years ago
15 points!!! GEOMETRY help please!!<br> Questions attached!
dolphi86 [110]
3. Look at the picture.

We have the right angle triangle. We know the sum of measures of angles in triangle is equal 180°. Therefore:

x^o+90^o+43^o=180^o\\\\x^o+133^o=180^o\ \ \ |-133^o\\\\x^o=47^o

Answer:\ \boxed{47^o}

4.
Look at the picture.

Use Pythagorean theorem:

x^2+14^2=(10+x)^2\\\\x^2+196=10^2+2\cdot10\cdot x+x^2\ \ \ \ |-x^2\\\\196=100+20x\ \ \ |-100\\\\20x=96\ \ \ |:20\\\\x=4.8

Used:\ (a+b)^2=a^2+2ab+b^2

Answer:\ \boxed{4.8\ units}

5.
TRUE: 1; 2; 4

6.
We find a slope of the line OP:

m=\dfrac{y_2-y_1}{x_2-x_1}\\\\O(2;\ 6)\to x_1=2;\ y_1=6\\\\P(4;\ 3)\to x_2=4;\ y_2=3\\\\m=\dfrac{3-6}{4-2}=\dfrac{-3}{2}

We have: OP:\ y=-\dfrac{3}{2}x+b

Now, we must find the slope of the line perpendicular to the line OP.
We know:

k:y=m_1x+b;\ l:y=m_2x+c\\\\k\ \perp\ l\iff m_1m_2=-1

therefore

-\dfrac{3}{2}m_2=-1\ \ \ |\cdot\left(-\dfrac{2}{3}\right)\\\\m_2=\dfrac{2}{3}

So. We have the answer! :)

Answer:\ \boxed{y=\dfrac{2}{3}x+\dfrac{1}{3}}

6 0
3 years ago
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