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Sedaia [141]
3 years ago
7

Identify the variables, the constant, the coefficients, and the exponents in the algebraic

Mathematics
1 answer:
Naya [18.7K]3 years ago
4 0
Variables: x and y
Coefficients: 5 and 2
Constants: 4
Exponents: 2 (if 3y is being squared)
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Which pair shows equivalent expressions?
dolphi86 [110]
Answer: <span>-2x-10=-2(x+5)</span>
<span>
-2x-10

Taking out common factor -2

= -2( x + 5 )</span>
4 0
3 years ago
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Gelppppppppp meeeeeeee
Alja [10]

Answer:

3/5

Step-by-step explanation:

sin is opposite over hypotenuse

find hypotenuse first -> Hypotenuse = 100 m

(notice that it is a 3-4-5 triangle)

60/100 = 6/10 = 3/5

3 0
3 years ago
Find the center of a circle with the equation: x2 y2−32x−60y 1122=0 x 2 y 2 − 32 x − 60 y 1122 = 0
mixas84 [53]

The equation of a circle exists:

$(x-h)^2 + (y-k)^2 = r^2, where (h, k) be the center.

The center of the circle exists at (16, 30).

<h3>What is the equation of a circle?</h3>

Let, the equation of a circle exists:

$(x-h)^2 + (y-k)^2 = r^2, where (h, k) be the center.

We rewrite the equation and set them equal :

$(x-h)^2 + (y-k)^2 - r^2 = x^2+y^2- 32x - 60y +1122=0

$x^2 - 2hx + h^2 + y^2 - 2ky + k^2 - r^2 = x^2 + y^2 - 32x - 60y +1122 = 0

We solve for each coefficient meaning if the term on the LHS contains an x then its coefficient exists exactly as the one on the RHS containing the x or y.

-2hx = -32x

h = -32/-2

⇒ h = 16.

-2ky = -60y

k = -60/-2

⇒ k = 30.

The center of the circle exists at (16, 30).

To learn more about center of the circle refer to:

brainly.com/question/10633821

#SPJ4

7 0
1 year ago
The Pythagorean Theorem can only be used with which type of triangle?
valkas [14]
Right triangle because  a2<span> + b</span>2<span> = c</span><span>2</span>
4 0
3 years ago
The triangles are similar.<br> What is the value of x?<br> Enter your answer in the box.<br><br> X=
Aleks04 [339]

Answer:

x=16

Step-by-step explanation:

5times 4=20

3times 4=12

4times 4=16

4 0
2 years ago
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