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il63 [147K]
3 years ago
9

Write a quadratic equation with integer coefficients having the given numbers as solutions.

Mathematics
1 answer:
slava [35]3 years ago
7 0

9514 1404 393

Answer:

  x² -22 = 0

Step-by-step explanation:

The roots are opposites, so the equation is pretty simple.

  x = ±√22

  x² = 22 . . . . . square both sides

  x² -22 = 0 . . . . your quadratic equation in standard form

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Is 22/13 equivalent to 13/4?
Amiraneli [1.4K]

Answer:

yes

Step-by-step explanation:

22/13 =1.69≈3.2

13/4 = 3.2

4 0
2 years ago
In if GS ⊥ XJ and XJ⊥EA what is true about GS and EA
svlad2 [7]

Answer:

(GS) || (EA)

Step-by-step explanation:

When two lines are parallel, all four angles formed by the intersection of the two lines are right angles, meaning their angle measures are (90°). This means that alternate interior angles are congruent because all alternate interior angles measures equal (90°). Therefore, by the alternate interior angles converse theorem, lines (GS) and (EA) are parallel.

The alternate interior angles converse theorem states if two angles are congruent and they have the relation of being a part of two lines intersected by a third line, then the two non-intersecting lines are parallel.

6 0
3 years ago
Read 2 more answers
Find the number b such that the line y = b divides the region bounded by the curves y = 36x2 and y = 25 into two regions with eq
Gemiola [76]

Answer:

b = 15.75

Step-by-step explanation:

Lets find the interception points of the curves

36 x² = 25

x² = 25/36 = 0.69444

|x| = √(25/36) = 5/6

thus the interception points are 5/6 and -5/6. By evaluating in 0, we can conclude that the curve y=25 is above the other curve and b should be between 0 and 25 (note that 0 is the smallest value of 36 x²).

The area of the bounded region is given by the integral

\int\limits^{5/6}_{-5/6} {(25-36 \, x^2)} \, dx = (25x - 12 \, x^3)\, |_{x=-5/6}^{x=5/6} = 25*5/6 - 12*(5/6)^3 - (25*(-5/6) - 12*(-5/6)^3) = 250/9

The whole region has an area of 250/9. We need b such as the area of the region below the curve y =b and above y=36x^2 is 125/9. The region would be bounded by the points z and -z, for certain z (this is for the symmetry). Also for the symmetry, this region can be splitted into 2 regions with equal area: between -z and 0, and between 0 and z. The area between 0 and z should be 125/18. Note that 36 z² = b, then z = √b/6.

125/18 = \int\limits^{\sqrt{b}/6}_0 {(b - 36 \, x^2)} \, dx = (bx - 12 \, x^3)\, |_{x = 0}^{x=\sqrt{b}/6} = b^{1.5}/6 - b^{1.5}/18 = b^{1.5}/9

125/18 = b^{1.5}/9

b = (62.5²)^{1/3} = 15.75

8 0
3 years ago
Plz save me I need help as sonic
Rzqust [24]

Answer:

1.Yes

2.Yes

3.No

4.No

5.Yes

Step-by-step explanation:

Just plug in the values of x and y for each set of coordinates.

Example from the first one

4x=y

4(5)=20

20=20 True

make sure all are valid by plugging in the next set of coordinates

4(6)=24

24=24 True

4(7)=28

28=28 True

and so forth with each equation plug in the coordinates

3 0
3 years ago
Read 2 more answers
Drag the labels to the correct locations on the table. Each label can be used more than once.
Julli [10]

Answer:

first one linear

second one quadratic

linear

Step-by-step explanation:

6 0
3 years ago
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