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4 years ago

12

Solve for x

- 1.5 = 0.23" alt="2 \sin(x) - 1.5 = 0.23" align="absmiddle" class="latex-formula">

1 answer:

zhenek [66]4 years ago

3 0

Add 1.5 to both sides:

$2\sin(x) = 1.83$

Divide both sides by 2:

$\sin(x) = 0.915$

Consider the arcsin of both sides:

$\arcsin(\sin(x)) = \arcsin(0.915)$

Arcsin and sin simplify, because they are inverse of each other:

$x = \arcsin(0.915)$

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Help me please i need help on my maths hw people

Nookie1986 [14]

65-14=65-5-x
x=9

Explanation=
65-14=65-5-x
51=60-x
-9=-x
9=x

6 0

3 years ago

Read 2 more answers

group of 30 students from your school is part of the audience for a TV game show the total number of people in the audience is 1

Soloha48 [4]

<h2>Answer:</h2><h2>The theoretical probability of four students from your school being selected as contestants out of 8 possible contestants spots = $\frac{14}{165}$ </h2>

Step-by-step explanation:

The number of students participated = 30

Total audience = 150

By probability , to find the solution = $\frac{n(E)}{n(S)}$

where n(E) is the number of favorable outcomes,

n(S) is the number of total outcomes.

n(S) is the number of ways any 8 students can by picked from the audience= $\frac{8}{140}$

n(E) is the probability of picking four students from our school and five students from another school.

n(E) = $(\frac{4}{30}) (\frac{4}{110})$ = $\frac{16}{3300}$

$\frac{n(E)}{n(S)}$ = $\frac{\frac{16}{3300}}{\frac{8}{140}}$ = $\frac{14}{165}$

7 0

3 years ago

Work out (5x10³) x (9x10)

nasty-shy [4]

Answer:

4.5 × 10^5 or 450000 :)))))

6 0

3 years ago

Find the directional derivative of at the point (1, 3) in the direction toward the point (3, 1). g

Anika [276]

Complete Question:

Find the directional derivative of g(x,y) = $x^2y^5$ at the point (1, 3) in the direction toward the point (3, 1)

Answer:

Directional derivative at point (1,3), $D_ug(1,3) = \frac{162}{\sqrt{8} }$

Step-by-step explanation:

Get $g'_x$ and $g'_y$ at the point (1, 3)

g(x,y) = $x^2y^5$

$g'_x = 2xy^5\\g'_x|(1,3)= 2*1*3^5\\g'_x|(1,3) = 486$

$g'_y = 5x^2y^4\\g'_y|(1,3)= 5*1^2* 3^4\\g'_y|(1,3)= 405$

Let P = (1, 3) and Q = (3, 1)

Find the unit vector of PQ,

$u = \frac{\bar{PQ}}{|\bar{PQ}|} \\\bar{PQ} = (3-1, 1-3) = (2, -2)\\{|\bar{PQ}| = \sqrt{2^2 + (-2)^2}\\$

$|\bar{PQ}| = \sqrt{8}$

The unit vector is therefore:

$u = \frac{(2, -2)}{\sqrt{8} } \\u_1 = \frac{2}{\sqrt{8} } \\u_2 = \frac{-2}{\sqrt{8} }$

The directional derivative of g is given by the equation:

$D_ug(1,3) = g'_x(1,3)u_1 + g'_y(1,3)u_2\\D_ug(1,3) = (486*\frac{2}{\sqrt{8} } ) + (405*\frac{-2}{\sqrt{8} } )\\D_ug(1,3) = (\frac{972}{\sqrt{8} } ) + (\frac{-810}{\sqrt{8} } )\\D_ug(1,3) = \frac{162}{\sqrt{8} }$

8 0

3 years ago

Need help and don’t just give answer have a explanation at least then i’ll give u brain list

juin [17]

Answer:

11.75

Step-by-step explanation:

Interior angles in a quadrilateral sum to 360.

Opposite angles in a parallelogram are the same.

360=118+118+(4x+15)+(4x+15)

= 236 + 8x + 30

= 266 + 8x

8x = 360 - 266 = 94

x=94/8 = 11.75

5 0

3 years ago

Read 2 more answers