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VikaD [51]
3 years ago
11

What are the solutions of x^2+10+16=0

Mathematics
1 answer:
goldenfox [79]3 years ago
8 0
-2 and -8 or (x+2),(x+8)
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What is the slope of a line perpendicular to 3x- 2y= 6
Tema [17]

Answer:

y = (-2/3)x + 5

2y=3x-6 Then divide by 2. y=3/2x-6 Now it is in slope-intercept form. The slope for this particular equation is 3/2 (m). The slope of a line that is perpendicular to this would be -2/3, the negative reciprocal of 3/2.

Step-by-Step:

HOPE THIS HELPED!!!

5 0
3 years ago
Read 2 more answers
Find the degree of the polynomial<br> 5x^3+7x^2-9
Fantom [35]

Answer:

The degree of the polynomial is 3.

6 0
2 years ago
Ms. Cruz has a tank with 11 fish. After buying two more, she has 2 less than triple the amount of fish Ms. Montesinos has. How m
notka56 [123]

Answer:

a) The variable, M represents the amount of fish Ms. Montesinos has

b) The set equation is 11 + 2 = 3 × M - 2

c) 5

Step-by-step explanation:

The given parameters are;

The number of fishes in Ms. Cruz's tank = 11 fishes

The number of fishes Ms, Cruz has after buying two more fishes = 3 × The amount of fish Ms. Montesinos has - 2 fishes

a) Let the variable, M represent the amount of fish Ms. Montesinos has

b) The set up equation is as follows;

11 + 2 = 3 × M - 2

c) From the equation, 11 + 2 = 3 × M - 2, we have;

13 = 3·M - 2

3·M = 13 + 2 = 15

3·M = 15

M = 15/3 = 5

M = 5

The number of fishes Ms. Montesinos has is 17 fishes

5 0
3 years ago
Geometry help. Inscribing circle help and constructing lines.
prohojiy [21]
<h3>1. Inscribe a circle ΔJKL. Identify the point of concurrency that is the center of the circle you drew. </h3>

To inscribe a circle in the attached triangle, we must follow the following steps. The resultant figure is the first one below.

1. Draw the angle bisector of angle JKL. The straight line in the green one.

2. Draw the angle bisector of angle KLJ. The straight line in the blue one.

3. Draw point D at these two lines.

4. Draw a line through point D perpendicular to side JK. The straight line in the red one.

5. Mark point G at the red line and side JK.

6. Draw a circle with center at D (point of concurrency) and a radius of DG.


<h3>2. Circumscribe a circle ΔFGH. Identify the point of concurrency that is the center of the circle you drew. </h3>

To circumscribe a circle in the attached triangle, we must follow the following steps:

1. Draw the perpendicular bisector of the side FG. The straight line in the red one. The resultant figure is the second one below.

2. Draw the perpendicular bisector of the side GH. The straight line in the blue one.

3. The center of the circle is the intersection of these two lines

4. Draw a circle with center at D (point of concurrency).

<h3>3. Construct the two lines tangents to circle</h3>

A tangent to a circle is a straight line that touches the circle at an only point. If a point A is outside a circle, we can draw two lines that pass through this point and are tangent to the circle at two different points each. So this is shown in the third figure. Lines red and blue are tangent to the circle at two different points.

3 0
3 years ago
5p-3r=1
Brilliant_brown [7]
If you would like to solve the system of equations, you can do this using the following steps:

5p - 3r = 1     /*2
8p + 6r = 4
__________
10p - 6r = 2
<span>8p + 6r = 4
</span>__________
10p - 6r + 8p + 6r = 2 + 4
18p = 6
p = 6/18
p = 1/3

<span>5p - 3r = 1
</span>5 * 1/3 - 3r = 1
5/3 - 3r = 1
5/3 - 1 = 3r
5/3 - 3/3 = 3r
2/3 = 3r
r = 2/9

(p, r) = (1/3, 2/9)

The correct result would be <span>(1/3, 2/9)</span>.
4 0
3 years ago
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