T

Mathematics

1 answer:

Rus_ich [418]2 years ago

4 0

Answer:

Step-by-step explanation:

yax2;

12+34+5+6=21 so y2+35

You might be interested in

An architect wants to draw a rectangle with a diagonal of 13 centimeters. The length of the rectangle is to be 3 centimeters mor

butalik [34]

Answer:

The rectangle is 3.2 cm by 12.6 cm

Step-by-step explanation:

See attached image for a diagram.

Choose w to represent the width because the length is described by referring to the width: it's 3 more than (add 3) triple (multiplied by 3) the width.

Length = 3w + 3

The diagonal forms two right triangles, each with leg = w, other leg = 3w + 3, hypotenuse = 13.

The Pythagorean Theorem says $(\text{leg})^2+(\text{leg})^2=(\text{hypotenuse})^2$ so

$w^2+(3w+3)^2=13^2\\\\w^2+9w^2+18w+9=169\\\\10w^2+18w-160=0\\\\5w^2+9w-80=0\\$

Now solve using the Quadratic Formula with $a=5,\,b=9,\,c=-80$ .

$w=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\w=\frac{-9\pm\sqrt{81+1600}}{10}\\\\w=\frac{-9\pm\sqrt{1681}}{10}\\\\w=\frac{-9\pm41}{10}\\\\w=-5\text{ or }w=3.2$

The negative root makes no sense as a distance, so the width of the rectangle is 3.2 cm. The length is 3(3.2) + 3 = 12.6 cm.

5 0

2 years ago

Solve each equation. List any restrictions. Check your solutions. 12/x+11=17

Thepotemich [5.8K]

12+11x=17x11x+12=17x
(both sides of the equation)
11x+12−17x=17x−17x
(Subtract 17x from both sides)
−6x+12=0−6x+12−12=0−12
(Subtract 12 from both sides)
−6x=−12−6x−6=−12−6
(Divide both sides by -6)
x=2
Check answers. (Plug them in to make sure they work.)
x=2
(Works in original equation)

6 0

2 years ago

Look at Figure X diagrammed on the graph below. Which of the following statements are true

zzz [600]

Check the picture below.

so, if we slap the bottom surplus to the empty area on the top, we can pretty much close off those squares above, that gives a solid area of at least a 4x5 square, the red part and the yellow part there, and that'd give us a perimeter of 4+4+5+5 = 20 at least.

we might also be able to augment the left-right sides together to close off those squares as well, but even if we did, our area will just be about a 4x6 or 24 square units.

A) is the area larger than 14 square units? ✔

B) is the area less than 35 square units? ✔

C) is the perimeter more than 16? ✔