Answer:
There will be 4 blue marbles
Step-by-step explanation:
Yellow to blue marbles is in a ratio of:
5:10
the representative sample is:
2:x
simplee cross multiplication can be aplied
5x = 2*10
5x = 20
divide both sides by 5
x = 4
there should be four blue marbles
Answer:
A. 2·x² + 16·x + 32 ≥ 254
Step-by-step explanation:
The given dimensional relationship between the dimensions of the photo in the center of the cake and the dimensions of the cake are
The width of the cake = The width of the photo at the center of the cake, x + 4 inches
The length of the cake = 2 × The width of the cake
The area of the cake Wanda is working on ≥ 254 in.²
Where 'x' represents the width of the photo (at the center of the cake), let 'W' represent the width of the cake, let 'L' represent the length of the cake, we get;
W = x + 4
L = 2 × W
Area of the cake, A = W × L ≥ 254
∴ A = (x + 4) × 2 × (x + 4) = 2·x² + 16·x + 32 ≥ 254
The inequality representing the solution is therefore;
2·x² + 16·x + 32 ≥ 254
Answer:
Not similar
Step-by-step explanation:
The measure of the angle ∠PQR is 90 degrees
<h3>How to prove that ∠PQR is 90 degrees?</h3>
The equation of the line PQ is given as:
3x - y - 2 = 0
The coordinates of the QR are given as:
(0, -2) and (6, -4)
Make y the subject in 3x - y - 2 = 0
y = 3x - 2
The slope of the above line is
m1 = 3
Next, we calculate the slope (m2) of points Q and R.
So, we have:
m2 = (y2- y1)/(x2 - x1)
This gives
m2 = (-4 + 2)/(6 - 0)
Evaluate
m2 = -1/3
The slopes of perpendicular lines are opposite reciprocals.
m1 = 3 and m2 = -1/3 are opposite reciprocals.
This means the lines PQ and QR are perpendicular lines.
The angle at the point of perpendicularity is 90 degrees
Hence, the measure of the angle ∠PQR is 90 degrees
Read more about linear equations at:
brainly.com/question/15602982
#SPJ1
Answer:
(c) (12th root of 8)^x
Step-by-step explanation:
The applicable rules of exponents are ...
![(a^b)^c=a^{bc}\\\\\sqrt[n]{a}=a^{1 \! /n}](https://tex.z-dn.net/?f=%28a%5Eb%29%5Ec%3Da%5E%7Bbc%7D%5C%5C%5C%5C%5Csqrt%5Bn%5D%7Ba%7D%3Da%5E%7B1%20%5C%21%20%2Fn%7D)
In this case, we have ...
![(\sqrt[3]{8})^{x/4}=8^{1/3\cdot x/4}=8^{x/12}=\boxed{\left(\!\sqrt[12]{8}\right)^x}](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7B8%7D%29%5E%7Bx%2F4%7D%3D8%5E%7B1%2F3%5Ccdot%20x%2F4%7D%3D8%5E%7Bx%2F12%7D%3D%5Cboxed%7B%5Cleft%28%5C%21%5Csqrt%5B12%5D%7B8%7D%5Cright%29%5Ex%7D)
