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Elenna [48]
2 years ago
15

A. 8 b. 4 c. 16 d. 12

Mathematics
2 answers:
lesya [120]2 years ago
6 0

3x =24

or, x= 24/3

or, x =8

Oxana [17]2 years ago
5 0

b. 4

....................

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Match each equation with its solution set.
taurus [48]

Answer:

1- The solution of I2x + 5I = 9 is {-7 , 2}

2- The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- The solution of I5 - xI = 6 is {-1 , 11}

4- The solution of I6x - 8I + 7 = 5 is ∅

5- The solution of Ix + 3I = 12 is {-15 , 9}

6- The solution of Ix - 3I = -12 is ∅

Step-by-step explanation:

* At first lets explain the meaning of IxI = a

- If IxI = a ⇒ then x = a or x = -a

- IxI never give a negative answer, because IxI means the

 magnitude of x is always positive

Ex: I-2I is 2

* Now lets find the solution of each equation

1- ∵ I2x + 5I = 9

∴ 2x + 5 = 9 ⇒ subtract 5 from both sides

∴ 2x = 4 ⇒ divide both sides by 2

∴ x = 2

OR

∴ 2x + 5 = -9 ⇒ subtract 5 from both sides

∴ 2x = -14 ⇒ divide both sides by 2

∴ x = -7

* The solution of I2x + 5I = 9 is {-7 , 2}

2- ∵ I2x + 7I + 2 = 11 ⇒ Subtract 2 from both sides

∴ I2x + 7I = 9

∴ 2x + 7 = 9 ⇒ subtract 7 from both sides

∴ 2x = 2 ⇒ divide both sides by 2

∴ x = 1

OR

∴ 2x + 7 = -9 ⇒ subtract 7 from both sides

∴ 2x = -16 ⇒ divide both sides by 2

∴ x = -8

* The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- ∵ I5 - xI = 6

∴ 5 -x = 6 ⇒ subtract 5 from both sides

∴ -x = 1 ⇒ divide both sides by -1

∴ x = -1

OR

∴ 5 -x = -6 ⇒ subtract 5 from both sides

∴ -x = -11 ⇒ divide both sides by -1

∴ x = 11

* The solution of I5 - xI = 6 is {-1 , 11}

4- ∵ I6x - 8I + 7 = 5 ⇒ Subtract 7 from both sides

∴ I6x - 8I = -2

- I  I never give negative answer

* The solution of I6x - 8I + 7 = 5 is ∅

5- ∵ Ix + 3I = 12

∴ x + 3 = 12 ⇒ subtract 3 from both sides

∴ x = 9

OR

∴ x + 3 = -12 ⇒ subtract 3 from both sides

∴ x = -15

* The solution of Ix + 3I = 12 is {-15 , 9}

6- ∵ Ix - 3I = -12

- I  I never give negative answer

* The solution of Ix - 3I = -12 is ∅

4 0
3 years ago
Both circle Q and circle R have a central angle measuring 75°. The ratio of circle Q's radius to circle R's radius is 2:5. Which
ycow [4]
The ratio of similar areas is the square of the ratio of the scale factor.

Circle R's sector is (5/2)² = 25/4 the area of Circle Q's sector.
7 0
3 years ago
Read 2 more answers
Help me please !!!!!!!! I DON'T KNOWWWWWWWWWW
r-ruslan [8.4K]

Answer:

m<1= 120

Step-by-step explanation:

1. we know supplementary angles equals 180. the angle 130 is part of a supplementary angle so we can subtract it by 180 to see that missing angle.

2. we also know that all triangles three angles add up to 180. So if we add the number we know 70+50 (50 is from subtracting 130 from 180), Once we find that answer, we subtract it from 180 to find angle 2 which is 60 degrees

3. Last step, angle 2 is part os a supplementary angle in which the two angles add up to 180. So, subtracting 60 from 180 will get you the degree of the angel one which his 120 degrees.

7 0
3 years ago
One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p
scZoUnD [109]

Answer:

c^2 = 9dp

Step-by-step explanation:

Given

dx^2 + cx + p = 0

Let the roots be \alpha and \beta

So:

\alpha = 2\beta

Required

Determine the relationship between d, c and p

dx^2 + cx + p = 0

Divide through by d

\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0

x^2 + \frac{c}{d}x + \frac{p}{d} = 0

A quadratic equation has the form:

x^2 - (\alpha + \beta)x + \alpha \beta = 0

So:

x^2 - (2\beta+ \beta)x + \beta*\beta = 0

x^2 - (3\beta)x + \beta^2 = 0

So, we have:

\frac{c}{d} = -3\beta -- (1)

and

\frac{p}{d} = \beta^2 -- (2)

Make \beta the subject in (1)

\frac{c}{d} = -3\beta

\beta = -\frac{c}{3d}

Substitute \beta = -\frac{c}{3d} in (2)

\frac{p}{d} = (-\frac{c}{3d})^2

\frac{p}{d} = \frac{c^2}{9d^2}

Multiply both sides by d

d * \frac{p}{d} = \frac{c^2}{9d^2}*d

p = \frac{c^2}{9d}

Cross Multiply

9dp = c^2

or

c^2 = 9dp

Hence, the relationship between d, c and p is: c^2 = 9dp

8 0
3 years ago
Which row of the input/output table is incorrect?
zaharov [31]

Row A:   x = 3

y = 2x - 3

y = 2(3) - 3 = 3      This is correct

Row B: x = 5

y = 2x - 3

y = 2(5) - 3 = 7        This is correct

Row C: x = 7

y = 2x - 3

y = 2(7) - 3 = 11      this is correct

Row D: x = 10

y = 2x - 3

y = 2(10) - 3 = 17  THIS IS INCORRECT, SO ROW D

8 0
3 years ago
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