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olasank [31]
2 years ago
15

Using the following image, solve for SR. Look at the image closely.

Mathematics
2 answers:
iren [92.7K]2 years ago
4 0

Step-by-step explanation:

I assume

SR = 2x + 23

RQ = x + 21

if that is true, then the situation is completely simple :

14 = (2x + 23) + (x + 21) = 3x + 44

3x = -30

x = -10

SR = 2×-10 + 23 = -20 + 23 = 3

RQ = -10 + 21 = 11

Vilka [71]2 years ago
4 0

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \:x = -10

____________________________________

\large \tt Solution  \: :

\qquad \tt \rightarrow \: SR + RQ = SQ

\qquad \tt \rightarrow \: 2x + 23 + x + 21 = 14

\qquad \tt \rightarrow \: 3x + 44 = 14

\qquad \tt \rightarrow \: 3x = 14 - 44

\qquad \tt \rightarrow \: 3x =  - 30

\qquad \tt \rightarrow \: x =  - 10

Now,

\qquad \tt \rightarrow \: SR = 2x + 23

\qquad \tt \rightarrow \: SR = 2(-10) + 23

\qquad \tt \rightarrow \: SR =-20+ 23

\qquad \tt \rightarrow \: SR = 3 \:\: units

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

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10x – 2 – 6x = 3x – 2 + x
mina [271]

Answer:

C

Step-by-step explanation:

4x-2=4x-2

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3 0
3 years ago
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A computer room has 12 computers. The room is open for 4 hours each day. 25 students sigh up for computer time. Each student get
GuDViN [60]

Answer:

155 minutes

Step-by-step explanation:

1. Find the total available "computer time": 4 hours * 12 computers= 48 hours

2. Split this among all of the students: 48 hours/25 students= 1.92 hours

3. Convert to minutes: 1.92 hr/student * 60 minutes ≈ 155 minutes/student

8 0
3 years ago
tom determines the system of equations below has two solutions, one of which is located at the vertex of the parabola. equation
stiks02 [169]
( x -3 )² = y - 4
y = ( x - 3 )² + 4
One solution is located at the vertex: ( 3, 4 ).
y = - x + b
4 = - 3 + b
b = 7
( x - 3 )² = - x + 7 - 4
x² - 6 x + 9 = - x + 3
x² - 5 x + 6 = 0
x² - 2 x - 3 x + 6 = 0
x ( x - 2 ) - 3 ( x - 2 ) = 0
( x - 2 ) ( x - 3 ) = 0
x 1 = 2,  x 2 = 3
y 1 = 5,  y 2 = 4.
In order for this solution to be reasonable, the 2nd equation must be:
y = - x + 7

8 0
3 years ago
X - 3y +3=0
Arte-miy333 [17]

Answer:

We know that for a line:

y = a*x + b

where a is the slope and b is the y-intercept.

Any line with a slope equal to -(1/a) will be perpendicular to the one above.

So here we start with the line:

3x + 4y + 5 = 0

let's rewrite this as:

4y = -3x - 5

y = -(3/4)*x - (5/4)

So a line perpendicular to this one, has a slope equal to:

- (-4/3) = (4/3)

So the perpendicular line will be something like:

y = (4/3)*x + c

We know that this line passes through the point (a, 3)

this means that, when x = a, y must be equal to 3.

Replacing these in the above line equation, we get:

3 = (4/3)*a + c

c = 3 - (4/3)*a

Then the equation for our line is:

y = (4/3)*x + 3 - (4/3)*a

We can rewrite this as:

y = (4/3)*(x -a) + 3

now we need to find the point where this line ( y = -(3/4)*x - (5/4)) and the original line intersect.

We can find this by solving:

(4/3)*(x -a) + 3 =  y = -(3/4)*x - (5/4)

(4/3)*(x -a) + 3  = -(3/4)*x - (5/4)

(4/3)*x - (3/4)*x = -(4/3)*a - 3 - (5/4)

(16/12)*x - (9/12)*x = -(4/3)*a - 12/4 - 5/4

(7/12)*x = -(4/13)*a - 17/4

x = (-(4/13)*a - 17/4)*(12/7) = - (48/91)*a - 51/7

And the y-value is given by inputin this in any of the two lines, for example with the first one we get:

y =  -(3/4)*(- (48/91)*a - 51/7) - (5/4)

  = (36/91)*a + (153/28) - 5/4

Then the intersection point is:

( - (48/91)*a - 51/7,  (36/91)*a + (153/28) - 5/4)

And we want that the distance between this point, and our original point (3, a) to be equal to 4.

Remember that the distance between two points (a, b) and (c, d) is:

distance = √( (a - c)^2 + (b - d)^2)

So here, the distance between (a, 3) and ( - (48/91)*a - 51/7,  (36/91)*a + (153/28) - 5/4) is 4

4 = √( (a + (48/91)*a + 51/7)^2 + (3 -  (36/91)*a + (153/28) - 5/4 )^2)

If we square both sides, we get:

4^2 = 16 =  (a + (48/91)*a + 51/7)^2 + (3 -  (36/91)*a - (153/28) + 5/4 )^2)

Now we need to solve this for a.

16 = (a*(1 + 48/91)  + 51/7)^2 + ( -(36/91)*a  + 3 - 5/4 + (153/28) )^2

16 = ( a*(139/91) + 51/7)^2 + ( -(36/91)*a  - (43/28) )^2

16 = a^2*(139/91)^2 + 2*a*(139/91)*51/7 + (51/7)^2 +  a^2*(36/91)^2 + 2*(36/91)*a*(43/28) + (43/28)^2

16 = a^2*( (139/91)^2 + (36/91)^2) + a*( 2*(139/91)*51/7 + 2*(36/91)*(43/28)) +  (51/7)^2 + (43/28)^2

At this point we can see that this is really messy, so let's start solving these fractions.

16 = (2.49)*a^2 + a*(23.47) + 55.44

0 = (2.49)*a^2 + a*(23.47) + 55.44 - 16

0 = (2.49)*a^2 + a*(23.47) + 39.44

Now we can use the Bhaskara's formula for quadratic equations, the two solutions will be:

a = \frac{-23.47  \pm  \sqrt{23.47^2 - 4*2.49*39.4}  }{2*2.49} \\\\a =  \frac{-23.47  \pm  12.57 }{4.98}

Then the two possible values of a are:

a = (-23.47 + 12.57)/4.98  = -2.19

a = (-23.47 - 12.57)/4.98 = -7.23

4 0
3 years ago
Johanna wrote the system of equations. 4 x - 3 y = 1, 5 x + 4 y = 9. If the second equation is multiplied by 4, what should the
polet [3.4K]

Answer:

The first equation must be multiplied by -5 to eliminate x variable by addition

Step-by-step explanation:

4 x - 3 y = 1 (1)

5 x + 4 y = 9 (2)

If the second equation is multiplied by 4

5x+4y=9. ×4

We have,

20x+16y=36 (3)

The first equation should be multiplied by -5 to eliminate x variable by addition

4x-3y=1 × -5

We have

-20x+15y=-5 (4)

Add equation (3) and (4) to eliminate x variable

20x+16y=36

-20x+15y=-5

31y=31

Divide both sides by 31

y=1

Substitute y=1 into equation (1)

4 x - 3 y = 1

4x-3(1)=1

4x-3=1

4x=1+3

4x=4

Divide both sides by 4

x=1

4 0
3 years ago
Read 2 more answers
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