Answer:
Step-by-step explanation:
Hello sorry I need points
The location of R on the number line will be 15/7.
Number line:
Number line is used for the visual representation of numbers on a straight line.
Basically, Zero (0) is considered to be the origin of a number line. The numbers to the left of 0 are negative numbers and the numbers to the right of 0 are all positive numbers.
Given,
On a number line,
point S is located at – 3 and
point T is located at 9.
Ratio of S and T = 3:4
We need to find the location of point R on S and T.
According to the given details,
The distance from S to T
=> 3 + 9 = 12
Through this we know that,
=> SR + RT = 12 ---------------------(1)
Based on the ratio,
S/T = 3/4
Which is similar to,
SR/RT = 3/4
So,
SR = 3/4 RT -----------(2)
Apply the value of SR on equation (1),
Then
3/4RT + RT = 12
=> 7/4 RT = 12
=> RT = 12 x (4/7)
=> RT = 48/7
Now the location of point R,
=> OT - RT = 9 - 48/7
=> 15/7
Hence "The location of R on the number line will be 15/7".
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Answer:

Step-by-step explanation:
since q(x)=-4
substitute
hence
-4=12x-3
take -3 to the other side where it becomes positive
therefore
-1=12x
divide by 12
you will get the above answer
Answer:
37.5 cm^2
Step-by-step explanation:
The area of a parallelogram is
A = bh where b is the base and h is the height
A = 7.5 * 5
A = 37.5 cm^2
Problem 1)
AC is only perpendicular to EF if angle ADE is 90 degrees
(angle ADE) + (angle DAE) + (angle AED) = 180
(angle ADE) + (44) + (48) = 180
(angle ADE) + 92 = 180
(angle ADE) + 92 - 92 = 180 - 92
angle ADE = 88
Since angle ADE is actually 88 degrees, we do NOT have a right angle so we do NOT have a right triangle
Triangle AED is acute (all 3 angles are less than 90 degrees)
So because angle ADE is NOT 90 degrees, this means
AC is NOT perpendicular to EF-------------------------------------------------------------
Problem 2)
a)
The center is (2,-3) The center is (h,k) and we can see that h = 2 and k = -3. It might help to write (x-2)^2+(y+3)^2 = 9 into (x-2)^2+(y-(-3))^2 = 3^3 then compare it to (x-h)^2 + (y-k)^2 = r^2
---------------------
b)
The radius is 3 and the diameter is 6From part a), we have (x-2)^2+(y-(-3))^2 = 3^3 matching (x-h)^2 + (y-k)^2 = r^2
where
h = 2
k = -3
r = 3
so, radius = r = 3
diameter = d = 2*r = 2*3 = 6
---------------------
c)
The graph is shown in the image attachment. It is a circle with center point C = (2,-3) and radius r = 3.
Some points on the circle are
A = (2, 0)
B = (5, -3)
D = (2, -6)
E = (-1, -3)
Note how the distance from the center C to some point on the circle, say point B, is 3 units. In other words segment BC = 3.