2. If AD= 2x – 1 and BC = 5x – 13, then AD= BC= AB= DC =

Mathematics

1 answer:

Sladkaya [172]3 years ago

4 0

Answer:

the question is missing a relation between AD AND BC

You might be interested in

In need of some help please.

Hatshy [7]

By the Pythagorean theorem
50^2 = x^2 +(x +34)^2
2500 = 2x^2 +68x + 1156
x^2 +34x -672 = 0
(x -14)(x +48) = 0
x = 14 or -48

The distance from the wall to the base of the ladder is 14 ft.

_____
7-24-25 is a Pythagorean triple. The dimensions here are double those values. It can be handy to know a few of the Pythagorean triples, as that can let you write down the answers to problems without having to go through the equations.

8 0

3 years ago

9) f(x) = x2 - X + 2<br> x = -8

timofeeve [1]

I think the answer will be 7.25.

5 0

4 years ago

The height h of a projectile is a function of the time t it is in the air. the height in feet for t seconds is given by the func

alexgriva [62]

Domain means the values of independent variable(input) which will give defined output to the function.

Given:

The height h of a projectile is a function of the time t it is in the air. The height in feet for t seconds is given by the function

$h(t)=-16t^2 + 96t$

Solution:

To get defined output, the height h(t) need to be greater than or equal to zero. We need to set up an inequality and solve it to find the domain values.

$To \; find \; domain:\\\\h(t) \geq0\\\\-16t^2+96t \geq 0\\Factoring \; -16t \; in \; the \; left \; side \; of \; the \; inequality\\\\-16t(t-6) \geq 0\\Step \; 1: Find \; Boundary \; Points \; by \; setting \; up \; above \; inequality \; to \; zero.\\\\t(t-6)=0\\Use \; zero \; factor \; property \; to \; solve\\\\t=0 \; (or) \; t = 6\\\\Step \; 2: \; List \; the \; possible \; solution \; interval \; using \; boundary \; points\\(- \infty,0], \; [0, 6], \& [6, \infty)$

$Step \; 3:Pick \; test \; point \; from \; each \; interval \; to \; check \; whether \\\; makes \; the \; inequality \; TRUE \; or \; FALSE\\\\When \; t = -1\\-16(-1)(-1-6) \geq 0\\-112 \geq 0 \; FALSE\\(-\infty, 0] \; is \; not \; solution\\Also \; Logically \; time \; t \; cannot \; be \; negative\\\\When \; t = 1\\-16(1)(1-6) \geq 0\\80 \geq 0 \; TRUE\\ \; [0, 6] \; is \; a \; solution\\\\When \; t = 7\\-16(7)(7-6) \geq 0\\-112 \geq 0 \; FALSE\\ \; [6, -\infty) \; is \; not \; solution$