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3 years ago

For 0° ≤ x < 360°, what are the solutions to sin(StartFraction x Over 2 EndFraction) + cos(x) – 1 = 0 ?

Mathematics

1 answer:

nataly862011 [7]3 years ago

6 0

Answer:

{60°,300°}

Step-by-step explanation:

edge 2020

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Alborosie

6578 i don’t know don’t ask me bit

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3 years ago

Identify the graphical solution to the inequality

NeTakaya

Answer:

Step-by-step explanation:

to solve absolute value inequalities you must solve it for when the absolute value is positive and then when it is negative

example, |4| = 4 and -4

first solution will be when we're solving for the positive value:

+(-4x+7) ≥ 27

-4x + 7 ≥ 27

-4x ≥ 20

x ≤ -5 [reverse the symbol when mult or div by a negative]

-(-4x+7) ≥ 27

4x - 7 ≥ 27

4x ≥ 34

x ≥ 34/4 or x ≥ 8 1/2

solution: x ≤ -5 or x ≥ 8.5

5 0

2 years ago

What is a reasonable estimate for the solution? O (1,-3/4) O (-3/4,1) O (-1,3/4) O (3/4,-1)

Zina [86]

Answer:

See Explanation

Step-by-step explanation:

Your question is incomplete, as the equations or graph or table(s) were not given.

However, I'll give a general way of solving this.

Take for instance, the equations are:

$y = \frac{4}{3}x - 1$

$y = \frac{2}{3}x - \frac{1}{2}$

To do this, we start by equating both equations.

$y = y$

i.e.

$\frac{4}{3}x - 1= \frac{2}{3}x - \frac{1}{2}$

Collect Like Terms

$\frac{4}{3}x - \frac{2}{3}x= 1 - \frac{1}{2}$

Take LCM

$\frac{4x- 2x}{3}= \frac{2 - 1}{2}$

$\frac{2x}{3}= \frac{1}{2}$

Cross Multiply

$2x * 2 = 3 * 1$

$4x = 3$

Make x the subject

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

Substitute 3/4 for x in $y = \frac{4}{3}x - 1$

$y = \frac{4}{3} * \frac{3}{4} - 1$

$y = 1 - 1$

$y = 0$

Hence:

$(x,y) = (\frac{3}{4},0)$

6 0

2 years ago

How do you do this math problem -4+7x+4=3

11111nata11111 [884]

Answer:

x = 3/7

x = 0.43

Step-by-step explanation:

Make x the subject

x > ×7 > + 4 > -4 = 3

3 > + 4 > -4 > ÷ 7 = x

7 0

3 years ago

In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= 2 cm.

konstantin123 [22]

Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.

1. Theorem: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

According to this theorem,

$BC^2=BD\cdot AB.$

Let BC=x cm, then AD=BC=x cm and BD=AB-AD=3-x cm. Then

$x^2=(3-x)\cdot 3,\\ \\x^2=9-3x,\\ \\x^2+3x-9=0,\\ \\D=3^2-4\cdot (-9)=9+36=45,\\ \\\sqrt{D}=\sqrt{45}=3\sqrt{5},\\ \\x_1=\dfrac{-3-3\sqrt{5} }{2}0.$

Take positive value x. You get

$AD=BC=\dfrac{-3+3\sqrt{5} }{2}\ cm.$

2. According to the previous theorem,

$AC^2=AD\cdot AB.$

Then

$AC^2=\dfrac{-3+3\sqrt{5} }{2}\cdot 3=\dfrac{-9+9\sqrt{5} }{2},\\ \\AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.$

Answer: $AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.$

This solution doesn't need CD=2 cm. Note that if AB=3cm and CD=2cm, then

$CD^2=AD\cdot DB,\\ \\2^2=AD\cdot (3-AD),\\ \\AD^2-3AD+4=0,\\ \\D$

This means that you cannot find solutions of this equation. Then CD≠2 cm.

8 0

3 years ago

Read 2 more answers