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timurjin [86]
3 years ago
12

Graph the line that represents this equation: y + 2 = (1+2) Drawing Tools Select Point Line​

Mathematics
1 answer:
Setler79 [48]3 years ago
5 0

Answer:

Look at the desmos graph below

Step-by-step explanation:

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5x + 2 + 3x + 5 = ?
Mama L [17]

Answer: 8x + 7 = ?

Step-by-step explanation: There isn't anything on the other side so all I can do is simplify the left.

8 0
2 years ago
HELP ASAP!!!!!!!!PLEASE<br><br> Solve for x. Please explain. Thank you
Stella [2.4K]

The values are x=8 and y=10

Explanation:

From the figure, we can see that the two quadrilaterals are similar.

Thus, by the property of quadrilateral, for similar quadrilaterals, their corresponding angles and sides are congruent. Thus, we have,

∠D = ∠S, ∠E = ∠P, ∠G = ∠R and ∠F = ∠Q

Also, GF = QR and DE = PS

Where QR = 2x-4 and ∠D = 102°, ∠G = 84° and ∠F = 68°, ∠Q = (6y+x)°

Hence, to determine the value of x for the side QR, we have,

Equating the values in GF = QR,we get,

12=2x-4\\16=2x\\8=x

Thus, the value of x is 8.

Substituting x in the angle Q to determine the value of y.

Thus, we have,

∠F = ∠Q

68=6y+x\\68=6y+8\\60=6y\\10=y

Thus, the value of y is 10.

Hence, The values are x=8 and y=10

8 0
2 years ago
The slope of the tangent line to the curve x^3y+y^2-x^2=5 at the point (2,1) is
zheka24 [161]

Answer:

-4/5

Step-by-step explanation:

To find the slope of the tangent to the equation at any point we must differentiate the equation.

x^3y+y^2-x^2=5

3x^2y+x^3y'+2yy'-2x=0

Gather terms with y' on one side and terms without on opposing side.

x^3y'+2yy'=2x-3x^2y

Factor left side

y'(x^3+2y)=2x-3x^2y

Divide both sides by (x^3+2y)

y'=(2x-3x^2y)/(x^3+2y)

y' is the slope any tangent to the given equation at point (x,y).

Plug in (2,1):

y'=(2(2)-3(2)^2(1))/((2)^3+2(1))

Simplify:

y'=(4-12)/(8+2)

y'=-8/10

y'=-4/5

5 0
2 years ago
8 x 1 5/12 = ????<br><br> Fraction
UNO [17]
For the answer is 10
4 0
2 years ago
Read 2 more answers
The slope of (17,2) and (18,-17)
Ahat [919]

<u>m= -19/1</u>

We need to use the slope equation

\frac{y_{2}-y_{1}  }{x_{2}-x_{1} }

We are working with the points,  

 (17,    2)         and      (18,     -17)

  x1    y1                       x2       y2

\frac{-17-2}{18-17}

<u>m= -19/1</u>

4 0
3 years ago
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