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3 years ago

Given sin x =.1234, find x in degrees.

1 answer:

7 0

You plug in the calculator sin−¹(.1234)
And it gives you 7.088
But since it asks you to round it your final answer would be:
7.1

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I don't understand how to do this, can someone please help me?

denis23 [38]

Answer:

x=1

Step-by-step explanation:

First find a common denominator between 7 and 4 which is 28, Multiply both equations by 28 eliminating the denominators so now you have theses two equations: 4(2x + 12) = 7(3x + 5) distribute these equations to get:

8x + 48 = 21x +35 now you want to get the x's on one side so 8-21 = -13 which as an equation so far would be -13x + 48 = 35 now do 35-48 = -13 so you can have 13x alone on one side. Your final equation now before you solve is

-13x = -13 the divide -13 by -13 to get 1.

Hope this helps you understand it better

4 0

3 years ago

Juliet went to the mall to buy clothes for the new school year. Shirts cost $15 and pants cost $40. If Juliet bought 10 items of

12345 [234]

7 shirts and 4 pants hope i helped

3 0

3 years ago

Read 2 more answers

What is the intersection of the three sets: A = {1, 3, 5, 7, 9}, B = {1, 2, 3, 5, 7}, and C = {1, 2, 5, 8, 9}?

marysya [2.9K]

ITS A. 1 AND 5 BECAUSE 1 AND 5 ARE IN ALL 3 SETS

7 0

3 years ago

Consider a total of 1,000 people. If it was said that within 1 deviation (of the mean) of all people like MATH 123, how many of

Lesechka [4]

Answer:

680 students

Step-by-step explanation:

The 68 - 95 - 99.7 rule (empirical rule) states that 68% of the population lies within one standard deviation of the mean, 95% of the population lies within two standard deviations and 95% of the population lies within three standard deviations.

Hence since it was said that within 1 deviation (of the mean) of all people like MATH 123, therefore the number of people that like MATH 123 is:

number of people that like MATH 123 = 68% of the population

number of people that like MATH 123 = 0.68 * 1000

number of people that like MATH 123 = 680 students

3 0

3 years ago

Find a solution x = x(t) of the equation x′ + 2x = t2 + 4t + 7 in the form of a quadratic function of t, that is, of the form x(

Temka [501]

The particular quadratic solution to the ODE is found as follows:

$x=at^2+bt+c$
$x'=2at+b$

$(2at+b)+2(at^2+bt+c)=t^2+4t+7$
$2at^2+(2a+2b)t+(b+2c)=t^2+4t+7$

$\begin{cases}2a=1\\2(a+b)=4\\b+2c=7\end{cases}\implies a=\dfrac12,b=\dfrac32,c=\dfrac{11}4$

Note that there's also the fundamental solution to account for, which is obtained from the characteristic equation for the ODE:

$x'+2x=0\implies r+2=0\implies r=-2$

so that $x_c=Ce^{-2t}$ is a characteristic solution to the ODE, and the general solution would be

$x=Ce^{-2t}+\dfrac{t^2}2+\dfrac{3t}2+\dfrac{11}4$

4 0

4 years ago