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Ipatiy [6.2K]
3 years ago
9

How do you write this in numbers thirty and fifteen-thousandths

Mathematics
1 answer:
Maru [420]3 years ago
7 0
I think the answer is 30.15. But I;m not positive. <span />
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Anna took a taxi from the airport to her house.
Gnom [1K]

Answer:

60-4=

56/2.80=

20 miles

Step-by-step explanation:

20 miles is your answer. Pls give brainliest

7 0
3 years ago
Slope intercept form
elena-14-01-66 [18.8K]
The slope intercept form is y = me+b
5 0
3 years ago
Which of the following constants can be added to x^2 + 2/3x to form a perfect square trinomial?
Angelina_Jolie [31]
Hello.

A trinomial is a perfect square if the square root of the first term times the square root of the third term times 2 equals the middle term.

\boxed{\mathsf{x^{2} + \dfrac{2}{3} x}}

a) Adding 1/9:

\cdot \: \mathsf{x^{2} + \dfrac{2}{3} x + \dfrac{1}{9}} \\ \\ \\ \mathsf{\sqrt{x^{2}} \times \sqrt{\dfrac{1}{9}} \times 2 =} \\ \\ \\ \mathsf{x \times \dfrac{1}{3} \times 2 =} \\ \\ \\ \mathsf{\dfrac{2}{3} x \rightarrow it \: is \: a \: perfect \: square \: trinomial}

b) Adding 4/9:

\cdot \: \mathsf{x^{2} + \dfrac{2}{3} x + \dfrac{4}{9}} \\ \\ \\ \mathsf{\sqrt{x^{2}} \times \sqrt{\dfrac{4}{9}} \times 2 =} \\ \\ \\ \mathsf{x \times \dfrac{2}{3} \times 2 =} \\ \\ \\ \mathsf{\dfrac{4}{3} x \rightarrow it \: is \: not \: a \: perfect \: square \: trinomial}

c) Adding 4 and 1/9:

\cdot \: \mathsf{x^{2} + \dfrac{2}{3} x + \dfrac{1}{9} + 4 = x^{2} + \dfrac{2}{3} x + \dfrac{37}{9}} \\ \\ \\ \mathsf{\sqrt{x^{2}} \times \sqrt{\dfrac{37}{9}} \times 2 =} \\ \\ \\ \mathsf{x \times \dfrac{\sqrt{37}}{3} \times 2 =} \\ \\ \\ \mathsf{\dfrac{2\sqrt{37}}{3} x \rightarrow it \: is \: not \: a \: perfect \: square \: trinomial}

Hope I helped.
3 0
3 years ago
el perimetro de un terreno rectangular es de 170m y su area es de 170^2hallar la medida de sus lados con ecuacion cuadratica
Blizzard [7]

Answer:

Acá tenemos un sistema de ecuaciones.

El perímetro de un rectángulo es:

2*A + 2*L = 170m

donde A es el ancho y L es el largo.

Y el área del rectángulo es:

A*L = 170m^2.

Entonces, el primer paso es aislar una de las variables en una de las ecuaciones, yo voy a aislar A en la segunda:

A = 170m^2/L.

Ahora reemplazo eso en la primera ecuación y la resuelvo para L.

2*( 170m^2/L.) + 2*L = 170m.

340m^2 + 2*L^2 = 170m*L

ahora tenemos una ecuación cuadrática:

2*L^2 - 170m*L + 340m^2 = 0.

Las soluciones se pueden obtener usando la formula de Bhaskara:

L = \frac{+170 +- \sqrt{170^2 -4*340*2}  }{2*2} = \frac{170 +-161.8}{4}

Entonces las soluciones son:

L = (170 + 161.8)/4 = 82.95m

L = (170 - 161.8)/4 = 4.1m

Entonces, si tomamos L = 82.95m, tenemos A = 4.1 m

y el área es:

A*L = 4.1m*82.95m = 170m^2

3 0
3 years ago
Find the product of (x + 3)(x − 3). <br><br> x2 − 6x + 9<br> x2 + 6x + 9<br> x2 + 9<br> x2 − 9
Scilla [17]
(x+3)(x-3)=
x^2-3x+3x-9=
x^2-9
8 0
3 years ago
Read 2 more answers
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