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15 days ago

Write three numbers for which you would use the hundreds place to compare to 35,712

Mathematics

1 answer:

Leona [4.1K]15 days ago

5 0

Answer:

35,694
35,981
35,102

Step-by-step explanation:

If you compare numbers based on the hundreds digit, the digits to the left (thousands and ten-thousands) remain the same, while the hundreds digit differs.

This applies for any five-digit numbers structured like:

(3)(5)(not 7)(any digit)(any digit).

Digits to the right of the decimal point, as well as those in the tens and ones places, don't impact this comparison.

You might be interested in

What values of b satisfy 3(2b + 3)2 = 36?

Inessa [3926]

To determine the values of b that fulfill 3(2b+3)^2 = 36

we start with
3(2b+3)^2 = 36
Divide both sides by 3
(2b+3)^2 = 12
Next, take the square root of both sides
(2b+3)} = (+ /-) \sqrt{12} \\ 2b=(+ /-) \sqrt{12}-3

b1=\frac{\sqrt{12}}{2} -\frac{3}{2}
b1=\sqrt{3} -\frac{3}{2}

b2=\frac{-\sqrt{12}}{2} -\frac{3}{2}
b2=-\sqrt{3} -\frac{3}{2}
Thus,

the solutions for b are
b1=\sqrt{3} -\frac{3}{2}
b2=-\sqrt{3} -\frac{3}{2}

9 0

3 days ago

Which statement is true about the discontinuities of the function f(x)? F (x) = StartFraction x minus 5 Over 3 x squared minus 1

AnnZ [3900]

Question:

Which statement is accurate concerning the function’s discontinuities $f(x) = \frac{x-5}{3x^2-17x-28}$

A) There are gaps at x = 7 and.

B) Asymptotes exist at x = 7 and.

C) Asymptotes exist at x = –7 and.

D) Gaps are present at (–7, 0) and.

Answer:

B) Asymptotes exist at x = 7 and $(x = \frac{-4}{3})$

Step-by-step explanation:

Given:

$f(x) = \frac{x-5}{3x^2-17x-28}$

Goal:

Identify the correct statement

$f(x) = \frac{x-5}{3x^2-17x-28}$

We need to factor the denominator first.

$f(x) = \frac{x-5}{(3x+4)(x-7)}$

To express x in terms of (3x+4) and (x-7):

3x + 4 =

3x = -4

Divide both sides by 3:

$x = \frac{-4}{3}$

x - 7

x = 7

Next, evaluate the limit when $(x = \frac{-4}{3})$ and at (x = 7)

lim f(x) as $(x = \frac{-4}{3})$ = ±∞

lim f(x) as (x=7) = ±∞

Since both scenarios result in the denominator approaching zero, they represent asymptotes.

Thus, asymptotes are found at $(x = \frac{-4}{3})$ and x=7

Option B is determined to be correct

6 0

3 days ago

Erika is writing a coordinate proof to show that the diagonals of a rectangle are congruent. She begins by assigning coordinates

Leona [4187]

Answer:

Option C is the right choice.

Step-by-step explanation:

The given coordinates define a rectangle, and our objective is to show that the diagonals JL and KM are congruent.

We know that rectangles possess four right angles.

To prove the congruence of the diagonals JL and KM, we will utilize the Pythagorean theorem.

In triangle KLM, KL has a length of b units while LM has a length of a units. By applying the Pythagorean theorem $\sqrt(a^{2}+b^{2})=KM$

In triangle JML, JM is b units long, and LM remains a units long. We again can apply the Pythagorean theorem $\sqrt(a^{2}+b^{2})=JL$

Thus, we find that $\sqrt(a^{2}+b^{2})=KM=LM$ and option C is the correct choice.

7 0

4 days ago

Read 2 more answers

Un globo vuela entre dos ciudades A y B, que distan entre sí 1.500 m. Los tripulantes del globo ven la ciudad A con un ángulo de

Inessa [3926]

Respuesta:

La altitud del globo por encima del nivel del suelo es de 449,6 metros.

Explicación paso a paso:

El enunciado no está completo. El texto completo es: "Un globo se desplaza entre las ciudades A y B, que se encuentran a 1.500 m de distancia. Los ocupantes del globo observan la ciudad A con un ángulo de depresión de 27°, mientras que, para la ciudad B, el ángulo es de 36°. ¿Cuál es la altura aproximada del globo con respecto al suelo?"

El diagrama que ilustra esta situación está en el archivo adjunto. Para calcular la altura del globo, se pueden emplear las funciones trigonométricas, siendo recomendable hacer uso de la función tangente para los ángulos de depresión mencionados:

Ciudad A

$\tan 27^{\circ} = \frac{h}{1500\,m-x}$