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stiv31 [10]
3 years ago
7

8x+2y=-2 y=-5x+1 Solve for x and y

Mathematics
2 answers:
Alexandra [31]3 years ago
8 0
That should be how you would solve the problem! Hope that helps:)

sp2606 [1]3 years ago
5 0
X = 2
y = -9
16 - negative 18 = -2

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Write the equation in slope-intercept form. -10x+2y=12<br><br> -please help
Goryan [66]

Answer:

y=5x+6

hope this helps

have a good day :)

Step-by-step explanation:

3 0
3 years ago
PLEASE HELP ME
olga_2 [115]

The measure of the angles are ∠A = 91°, ∠C = 89° and ∠D = 34°

Explanation:

Given that the quadrilateral ABCD is inscribed in a circle.

The given angles are ∠A = (2x + 3), ∠C  = (2x + 1) and ∠D = (x - 10)

We need to determine the measures of the angles A, C and D

<u>The value of x:</u>

We know that, the opposite angles of a cyclic quadrilateral add up to 180°

Thus, we have,

\angle A+\angle C=180^{\circ}

Substituting the values, we have,

2x+3+2x+1=180

             4x+4=180

                   4x=176

                     x=44

Thus, the value of x is 44.

<u>Measure of ∠A:</u>

Substituting x=44 in ∠A = (2x + 3)°, we get,

\angle A=(2(44)+3)^{\circ}

     =(88+3)^{\circ}

\angle A=91^{\circ}

Thus, the measure of angle A is 91°.

<u>Measure of ∠C :</u>

Substituting x=44 in ∠C = (2x + 1)°, we get,

\angle C=(2(44)+1)^{\circ}

     =(88+1)^{\circ}

\angle C=89^{\circ}

Thus, the measure of angle C is 89°.

<u>Measure of ∠D :</u>

Substituting x=44 in ∠D = (x - 10)°, we get,

\angle D=(44-10)^{\circ}

\angle D=34^{\circ}

Thus, the measure of angle D is 34°.

Hence, the measure of the angles are ∠A = 91°, ∠C = 89° and ∠D = 34°

7 0
3 years ago
8.
svp [43]
A 80 square meter I’m not quite sure
7 0
3 years ago
Determine whether the ordered pair (6,5) is a solution to the equation -2x + y = -7
natita [175]

Answer:

i really dont knok but my guess is yes

Step-by-step explanation:

8 0
3 years ago
(3cosx-4sinx) + (3sinx+4cosx) = 5
Ksenya-84 [330]
(3 cos x-4 sin x)+(3sin x+4 cos x)=5
(3cos x+4cos x)+(-4sin x+3 sin x)=5
7 cos x-sin x=5
7cos x=5+sin x
(7 cos x)²=(5+sinx)²
49 cos²x=25+10 sinx+sin²x
49(1-sin²x)=25+10 sinx+sin²x
49-49sin²x=25+10sinx+sin²x
50 sin² x+10sinx-24=0

Sin x=[-10⁺₋√(100+4800)]/100=(-10⁺₋70)/100
We have two possible solutions:
sinx =(-10-70)/100=-0.8
x=sin⁻¹ (-0.8)=-53.13º      (360º-53.13º=306.87)

sinx=(-10+70)/100=0.6
x=sin⁻¹ 0.6=36.87º

The solutions when  0≤x≤360º are:  36.87º and 306.87º.


3 0
3 years ago
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