The sum of two numbers is 41 . The smaller number is 11 less than the larger number . What are the numbers ?

Mathematics

1 answer:

AleksAgata [21]3 years ago

4 0

26 and 15 is the answer!

You might be interested in

The length of a rectangle is 5in longer than its width. If the perimeter of the rectangle is 46in, find its area

Crank

width = x
length = x+5

2x + 2(x+5) = 46
2x + 2x + 10 = 46
4x = 46 - 10
4x = 36
x = 36/4 
x = 9 in ← width

length = x+5 = 9 + 5 = 14 in

Area = 9 * 14 = 126 in²

6 0

3 years ago

Solve the equation for x∈Z -x² +8x -14 ≥ 0

kenny6666 [7]

$ax^2+bx+c=0\\\Delta=b^2-4ac\\if\ \Delta < 0\ then\ no\ solutions\\if\ \Delta=0\ then\ one\ solution:x=\frac{-b}{2a}\\if\ \Delta > 0\ then\ two\ solutions:x=\frac{-b\pm\sqrt\Delta}{2a}\\================================\\\\-x^2+8x-14\geq0\ \ \ \ |multiply\ both\ sides\ by\ (-1)\ \{change\ \geq\ on\ \leq\}\\\\x^2-8x+14\leq0\\\\a=1;\ b=-8;\ c=14\\\\\Delta=(-8)^2-4\cdot1\cdot14=64-56=8 > 0\\\sqrt\Delta=\sqrt8=\sqrt{4\cdot2}=\sqrt4\cdot\sqrt2=2\sqrt2$

$x_1=\frac{-(-8)-2\sqrt2}{2\cdot1}=\frac{8-2\sqrt2}{2}=\frac{8}{2}-\frac{2\sqrt2}{2}=\boxed{4-\sqrt2}\\\\x_2=\frac{-(-8)+2\sqrt2}{2\cdot1}=\frac{8+2\sqrt2}{2}=\frac{8}{2}+\frac{2\sqrt2}{2}=\boxed{4+\sqrt2}\\\\============================$

$ax^2+bx+c=0\\\\if\ a > 0\ then\ the\ parabola\ o pen\ up\\if\ a < 0\ then\ the\ parabola\ o pen\ down\\========================\\\\a=1 > 0-therefore\ o pen\ up\ (look\ at\ the\ picture)\\\\===============================\\\\Answer:x\in\left$

$Solutions\ in\ \mathbb{Z}:\\\\\sqrt2\approx1.4\\\\threfore:4-\sqrt2\approx4-1.4=2.6\ and\ 4+\sqrt2\approx4+1.4=5.4\\\\look\ at\ the\ second\ picture:x=3\ or\ x=4\ or\ x=5\ (x\in\{3;\ 4;\ 5\})$