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dusya [7]
3 years ago
7

8 increased by three times a number

Mathematics
1 answer:
nexus9112 [7]3 years ago
6 0
The answer would be 24
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2+x=15-7/8x how do you solve this?
uranmaximum [27]

Answer:2+x=15-7/8x

We move all terms to the left:

2+x-(15-7/8x)=0

Domain of the equation: 8x)!=0

x!=0/1

x!=0

x∈R

We add all the numbers together, and all the variables

x-(-7/8x+15)+2=0

We get rid of parentheses

x+7/8x-15+2=0

We multiply all the terms by the denominator

x*8x-15*8x+2*8x+7=0

Wy multiply elements

8x^2-120x+16x+7=0

We add all the numbers together, and all the variables

8x^2-104x+7=0

a = 8; b = -104; c = +7;

Δ = b2-4ac

Δ = -1042-4·8·7

Δ = 10592

The delta value is higher than zero, so the equation has two solutions

We use following formulas to calculate our solutions:

x1=−b−Δ√2ax2=−b+Δ√2a

The end solution:

Δ−−√=10592−−−−−√=16∗662−−−−−−−√=16−−√∗662−−−√=4662−−−√

x1=−b−Δ√2a=−(−104)−4662√2∗8=104−4662√16

x2=−b+Δ√2a=−(−104)+4662√2∗8=104+4662√16

Step-by-step explanation:

5 0
3 years ago
How do I simply this, thanks!
stepan [7]
(-a^3b^2*-a^-2b^-3)^-2/2a^2b^-3 = a^4b^9/2a^8b^4 =b^5/2a^4 so your answer is b^5/2a^4
7 0
3 years ago
The vertices of a triangle are, A (2, 5) B (1, 2) C (3, 1),. Find the coordinates of the image after a
chubhunter [2.5K]

Answer:

The images are;

A’ (5,2) B’ (2,1) and C’ (1,3)

Step-by-step explanation:

Firstly, we want to reflect across the x-axis

When we reflect a point (x,y) over x-axis, we get (x,-y)

So;

A ( 2,5) becomes (2,-5)

B (1,2) becomes (1,-2)

C ( 3,1) becomes (3,-1)

Then we proceed to rotate 90 degrees counterclockwise about the origin

Here we have;

(x,y) becomes (-y,x)

(2,-5) becomes (5,2)

(1,-2) becomes (2,1)

(3,-1) becomes (1,3)

4 0
3 years ago
Help!! and it said 20 characters so....
Levart [38]

The volume is 1562.4 cm^3

V = (173.6 cm^2)(9cm)

   = 1562.4 cm^3

6 0
3 years ago
Plot the image of point D under a dilation about the origin (0,0) with a scale factor of 3.
babunello [35]

Answer:

D'(-6, 0) is the image of point D(-2, 0) under a dilation about the origin (0,0) with a scale factor of 3, as shown in attache figure below.

Please check the attache graph below where you can check the plot of the image D' of point D.

Step-by-step explanation:

As the point D on the coordinate plane has the vertex location

  • D(-2, 0)

The rule of dilation about the origin (0,0) with a scale factor of 3:

The rule states that if we dilate the point, let say D(-2, 0), under a dilation about the origin (0,0) with a scale factor of 3, all we need to multiple the coordinates of the point D with 3. The result will be the image of point D.

so

  • P(x, y)           →         P'(3x, 3y)
  • D(-2, 0)         →         D'(-6, 0)                ∵ (-2×3, 0×3)=(-6, 0)

Therefore, the image of point D under a dilation about the origin (0,0) with a scale factor of 3 can be plotted at the location D'(-6, 0).

So the new location of  D' is (-6, 0).

D'(-6, 0) is the image of point D(-2, 0) under a dilation about the origin (0,0) with a scale factor of 3, as shown in attache figure below.

Please check the attache graph below where you can check the plot of the image D' of point D.

3 0
3 years ago
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