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enot [183]
4 years ago
8

Solve the equation in the interval from 510^\circ510 ∘ 510, degrees to 1050^\circ1050 ∘ 1050, degrees. Your answer should be in

degrees. \sin(x)=0sin(x)=0
Mathematics
1 answer:
MakcuM [25]4 years ago
8 0

Answer:

<h2>540°, 720° and 900°</h2>

Step-by-step explanation:

Given equation sin(x) = 0 lying between the interval 510°≤x≤1050°. To find the value of x within this range, we first find the value of from the equation as shown;

sin(x) = 0

taking the arcsin of both sides;

sin^{-1}(sinx) =  sin^{-1} 0\\x = 0^{0}

Since x is positive in the first and second quadrant, in the second quadrant, the value of x will be 180°-0° = 180°

<em>Subsequent value of x equivalent to 0° will be addition of 180° to each previous value gotten</em>. The values of x within the range given are as shown

x1 =0°

x2 = 180°-0° = 180(2nd quadrant)

x3 = 180°+180° = 360°

x4 = 360°+180° = 540°

x5 = 540°+180°=720°

x6 = 720°+180° = 900°

x7 = 900°+180° = 1080°

We can see from the values above that the values of x that falls within the given range are 540°, 720° and 900°. This gives the required answers in degrees.

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Chris needs to make a 500 L of a 35% acidic solution. He has only two of the acidic solutions available, a 25% solution and a 50
DaniilM [7]
Yes, very fun, not


ok
one way is table or

solutions amount x and y
x+y=500
x=25%
y=50%
0.25x+0.5y=0.35(500)

we have
x+y=500
0.25x+0.5y=0.35(500)
0.25x+0.5y=175
solve

multiply second equation by 100
25x+50y=17500
divide everybody by 25
x+2y=700

now we have
x+y=500
x+2y=700

multiply first equation by -1 and add to second equaiton

-x-y=-500
<u>x+2y=700 +</u>
0x+1y=200

y=200
sub
x+y=500
x+200=500
x=300


300L of the 25% and 200L of the 50%
4 0
3 years ago
Solve this equation using an algebraic method: (x + 4)( x - 4) = 9<br> Help me out please
Artyom0805 [142]
You have to FOIL out the (x+4)(x-4) and then subrtract away the 9 in order to get a quadratic that you can solve for x.  As it is, you can't do it.
(x+4)(x-4)= x^{2} -4x+4x-16= x^{2} -16
Now if you move the 9 over with it, you get this:
x^{2} -16-9=0
which simplifies to
x^{2} -25=0
Now you can either solve this by recognizing that is the difference of perfect squares, or you can move the 25 over to the other side and take the square root of both sides, like this:
x^{2} =25
\sqrt{ x^{2} } = \sqrt{25}
x=+/-5
5 0
4 years ago
If the endpoints of the diameter of a circle are (10, 12) and (0, 2), what is the standard form equation of the circle? A) (x +
attashe74 [19]
So the equation of a circle is (x - h)² + (y - k)² = r² where (h,k) are the coordinates of the center of the circle and r is the radius. The diameter of a circle is a line that goes from one point of the circle to the other through the center of the circle. Well the center would be midway through the diameter so use midpoint formula to find the center which is (h,k) Mid point formula is both given x's added together divided by 2 for h and both y coordinates added together divided by 2 to find k
(10+0)/2
10/2= 5
(12+2)/2
14/2 = 7
so the center of the circle is (5,7) now use distance formula using the center and one of the points to the radius
√((5-10)²+(7-12)²)
√(-5²+ -5²)
√(25 + 25)
√50 is the radius
Now plug all found information into circle equation
(x-5)² + (y-7)² =50      note the end is 50 because the circle equation is radius squared and since the radius is √50, radius² is 50.
Answer is c
8 0
4 years ago
Read 2 more answers
Please find the missing side lengths
alex41 [277]
The missing side lengths
6)B.15
7)A.8
4 0
3 years ago
a quadrilateral ABCD in which angle DAB is a right angle. AB is 3.3cm, BC is 3.9cm, CD is 5.2cm and DA is 5.6cm. Find the length
Greeley [361]

Answer:

Part A) BD = 6.5 cm

Part B) see explanation

Step-by-step explanation:

See the attached figure

Part A) Find the length of BD

ΔDAB is a right triangle at A,  AB is 3.3cm ,  DA is 5.6cm

So, DB is the hypotenuse

Using Pythagorean equation:

DB = √(AD² + AB²) = √(3.3² + 5.6²) = √42.25 = 6.5 cm

===================================================

Part B) Show that angle BCD is 90°​

Given: CD is 5.2cm , BC is 3.9cm  and DB = 6.5 cm

So, CD² = 5.2² = 27.04

BC² = 3.9² = 15.21

DB² = 6.5² = 42.25

So, CD² + BC² = 27.04 + 15.21 = 42.25 = DB²

So, DB represent a hypotenuse at ΔBCD

So, the apposite angle of BD is a right angle

So, ∠BCD = 90°

5 0
3 years ago
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