(20 points, please help!) Which name is a correct way to name the tiling?

Mathematics

1 answer:

Naddika [18.5K]3 years ago

7 0

Answer:

Actually ignore what i put.. I counted again and realized there were actually 19..

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Guys please help me I’ll give u 7 points and mark brainliest need asap

enot [183]

Answer:

a. y = 3x + 4.5 & y = 7x + 2

b. x = 0.625 (year)

c. 6.375 feet

Explanation:

Type A starts at 4.5 and Type B starts at 2 ,each year can be marked as x

type A grow 3 inches per year = 3x

type B grows 7 inches per year = 7x

y = the total height

Since they both equal to y, you can set them equal to each other

3x + 4.5 = 7x +2

After solve you get answer for b

Work: 3x + 4.5 = 7x + 2

4.5 = 4x + 2

2.5 = 4x

0.625 = x

Substitute the value of x you got from part b into each equation then you will get 6.375 as your y value which is also your height

8 0

3 years ago

PLEASE HELP!!!!!! the sum of a number times 10 and 15 is at most -17

Brilliant_brown [7]

Let the number = x

Writing the equation you have:

10x + 15 <= -17

Solve for x

Subtract 15 from both sides:

10x <= -32

Divide both sides by 10:

X <= -32/10

Simplify:

X <= -16/5 as a fraction or -3.2 as a decimal

3 0

3 years ago

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

kenny6666 [7]

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

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

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

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