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MAVERICK [17]
4 years ago
10

cooghan asked his 19 classmates whether they were right or left-handed. there were 5 more right-handed classmates than left-hand

ed classmates
Mathematics
1 answer:
KIM [24]4 years ago
8 0
Divide 19 by 5 
and that should be it

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What are the intercepts of -4x+6y-5z=60
AURORKA [14]
Equation is <span>-4x+6y-5z=60
x-intercept is obtained by assuming y=0, z=0, and solving for x.
-4x+0-0=60 => x=-15, hence x-intercept is (-15,0,0)
Similarly for y-intercept, assume x=0,z=0 =>
0+6y-0=60 => y=60/6=10 => y-intercept is (0,10,0)
Again, for the z-intercept, assume x=0,y=0 =>
0+0-5z=60 => z=-12 => z-intercept is (0,0,-12)</span>
7 0
3 years ago
The design of a digital box camera maximizes the volume while keeping the sum of the dimensions at 6 inches. If the length must
love history [14]
H = x, L = 1.5 x;  
L + W + H = 6
x + 1.5 x + W = 6
W = 6 - 2.5 x
Dimensions of the camera in terms of x:
x,  1.5 x,  6 - 2.5 x.
V = L x W x H
V = x * 1.5 x * ( 6 - 2.5 x ) = 9 x² - 3.75 x³
V ` = 18 x - 11.25 x² ( V max is when V` = 0 )
18 x - 11.25 x² = 0
x ( 18 - 11.25 x ) = 0
11.25 x = 18
x = 18 : 11.25
x = 1.6
The dimensions are:
L = 2.4
W = 2.0
H = 1.6
4 0
3 years ago
Solve for x.<br><br> x/2= -5<br><br> x=5/2<br> x=-5/2<br> x=-10
NNADVOKAT [17]
X/2 × 2 = -5 × 2

x = -10
6 0
3 years ago
listed in the table is the percentage of students who choose each type of milk at lunchtime. use the table to determine the meas
sladkih [1.3K]

Answer:

i cant see the picture its blocked

Step-by-step explanation:

5 0
3 years ago
the length of a rectangle is 2 in more than it's w i d t h the area of a rectangle is equal to 1 inch less than three times a pe
Marizza181 [45]
If we let x and y represent length and width, respectively, then we can write equations according to the problem statement.
.. x = y +2
.. xy = 3(2(x +y)) -1

This can be solved a variety of ways. I find a graphing calculator provides an easy solution: (x, y) = (13, 11).

The length of the rectangle is 13 inches.
The width of the rectangle is 11 inches.

______
Just so you're aware, the problem statement is nonsensical. You cannot compare perimeter (inches) to area (square inches). You can compare their numerical values, but the units are different, so there is no direct comparison.

6 0
3 years ago
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