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3 years ago

Estimate the answers to the following calculations by carrying out order of magnitude calculations.

Mathematics

1 answer:

alina1380 [7]3 years ago

7 0

9514 1404 393

Answer:

a) 230

b) 300

c) 2×10^-2

d) 1.1×10^-6

Step-by-step explanation:

a) 0.26 × 890 ≈ 1/4 × 900 ≈ 225 ≈ 230

b) 1.95/(.67×10^-2) ≈ 2/(2/3)×10^2 ≈ 300

c) 2010×10^-5 ≈ 2×10^3×10^-5 = 2×10^-2

d) 9.98×10^-4/(9×10^2) = 9.98/9×10^(-4-2) ≈ 1.1×10^-6

YMMV depending on how you do the rounding and approximate multiplication and division.

The first one can be done multiple ways. For most accurate results, increasing one number while decreasing the other is recommended. (You don't want to compute 0.3×900, for example.)

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choli [55]

Answer:

Check below.

Step-by-step explanation:

1/3 = 0.33334

$\sqrt{0.08}$ = 0.24

-1.2

$-\sqrt{\frac{5}{3} }$ = -1.29

7 0

3 years ago

Read 2 more answers

Solve the system be either substitution or elimination 3x-5y=21 and 2x+y=1

IRISSAK [1]

Answer:

(2,-3)

Step-by-step explanation:

I used substitution and rearranged the second equation to y=-2x+1. Then I replaced the y in the first equation with -2x+1. So, I got 3x-5(-2x+1)=21. When you solve that you get 13x=26 or x=2. Then you replace x with 2 in either equation and end up with y=-3.

4 0

4 years ago

Read 2 more answers

Ayuda es para mañana El estudiante 1 y el estudiante 2 están a 53 metros, uno del otro. El estudiante 1, desde su posición, ve u

hram777 [196]

Answer:

La opción correcta es;

Estudiante 2, ya que está aproximadamente a 32,39 metros de distancia

Step-by-step explanation:

Los parámetros dados son;

La distancia entre los dos estudiantes = 53 metros

El ángulo medido entre la línea del Estudiante 1 a la caja fuerte y el Estudiante 1 y el Estudiante 2 = 37 °

El ángulo medido entre la línea del Estudiante 2 a la caja fuerte y el Estudiante 2 y el Estudiante 1 = 43 °

Dado que los dos estudiantes y la caja fuerte forman los vértices de un triángulo, tenemos;

Sea θ el ángulo del triángulo opuesto a la distancia entre los dos estudiantes

Por lo tanto;

θ + 43 ° + 37 ° = 180 ° (teorema de suma de ángulos)

θ = 180 ° - (43 ° + 37 °) = 100 °

Por regla del seno, tenemos;

a / (sin (A) = b / sin (B)

Por lo tanto;

53 / (sin (100 °)) = b / (sin (37 °)) = c / (sin (43 °))

Lo que da;

b = (sin (37 °)) × 53 / (sin (100 °)) = 32,39 metros

c = (sin (43 grados)) × 53 / (sin (100 grados)) = 36,7 metros

Por lo tanto, el Estudiante 2 está más cerca, ya que está aproximadamente a 32,39 metros de distancia.

6 0

3 years ago

Please help!

lidiya [134]

Let’s start out by breaking down the equation

1. eight more than. we know this must mean we are adding 8, so (+ 8)

2. the difference of a number and sixteen. we know the number (x) and 16 are being subtracting as indicated by the word difference, so (x-16)

now let’s combine them (x-16) + 8

let’s set it equal to 0, just to confirm it is correct.
(x-16) +8 = 0
x-16 = -8
x = 8

so substituting x for 8 in the translation: (8-16) + 8 = 0
-8 +8=0

it is correct! :)

6 0

4 years ago

The chorus has six sopranos and eight baritones. in how many ways can the director choose a quartet that contains at least one s

adell [148]

If you have six sopranos and eight baritones, you have fourteen musicians in total.
The number of possible quartets that you can get with 14 musician is:
$14C4= \frac{14!}{4!(14-4)!}$
$14C4= \frac{14!}{(4!)(10!)}$
$14C4=1001$

Now, the number of possible quartets without a soprano that you can have is:
$8C4= \frac{8!}{4!(8-4)!}$
$8C4= \frac{8!}{(4!)(4!)}$
$8C4=70$

Therefore, the number of quarters that contains at list one soprano will be the number of possible quarters minus the number of quarters without a soprano:
$1001-70=931$
We can conclude that the director can choose a quarter which contains at least one soprano in 931 ways.

4 0

3 years ago