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Tanya [424]
3 years ago
15

Solve the equation for the given variable 11=−4x+3

Mathematics
2 answers:
bekas [8.4K]3 years ago
6 0

Answer: x=-2

Step-by-step explanation:

Given

11=-4x+3

Subtract 3 on both sides

11-3=-4x+3-3

8=-4x

Divide -4 on both sides

-4x/-4=8/-4

<h2>x=-2</h2>
Diano4ka-milaya [45]3 years ago
6 0

Answer: x=-2

Step-by-step explanation:

Subtract 3 from both sides

1) 8= -4x

2) Isolate the variable (divide -4 on both sides)

3) X=-2

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11/(x+9)=5<br> I am struggling with this question. Is there a solution?
maks197457 [2]

Answer:

x=-6.8

\\ step \: by \: step \: explanation \\  \frac{11}{(x + 9)}  = 5 \\ 11 = 5(x + 9) \\ 11 = 5x + 45 \\ 5x = 11 - 45 \\ 5x =  - 34 \\ x =   \frac{ - 34}{5}  =  - 6.8

8 0
2 years ago
(x+6)(x^2-7x+12)<br> Algebra addition problem to find the area.
lisov135 [29]

Answer:

=x^3-x^2-30x+72

Step-by-step explanation:

Distribute parentheses:

=xx^2+x\left(-7x\right)+x\cdot \:12+6x^2+6\left(-7x\right)+6\cdot \:12

Apply minus - plus rule:

=x^2x-7xx+12x+6x^2-6\cdot \:7x+6\cdot \:12

Simplified:

=x^3-x^2-30x+72

Its complicated but..

7 0
3 years ago
Read 2 more answers
Write the equation for the line tangent to the circle x2 + y2 = 29 at the point (2, 5).
oee [108]
X2 + y2 = 29

Center of the circle 
x=0 and y=0

Therefore the slope of the line perpendicular to the tangent line
(y2-y1) / (x2-x1)
(5-0) / (2-0)
5/2
 
 so the slope of the tangent 
 -2/5             

General equation for the line
y = mx + c
if the line satisfies with the point (2,5)
5 = 2m + c

m = -2/5

5 = 2 * (-2/5) + c
c = 5 + (4/5)
c = 29/5

Therefore the equation of the line
y = mx + c
y = (-2/5) m + 29/5 
3 0
3 years ago
13) Mark walks 3.4 miles per day. How<br> many days will it take him to walk 17.34<br> miles?
HACTEHA [7]

Answer:

5.1 days

Step-by-step explanation:

3.4 \: miles = 1 \: day \\ 17.34 \: miles \:  = x \: days

Cross Multiply

3.4x = 17.34

Divide both sides of the equation by 3.4

\frac{3.4x}{3.4}  =  \frac{17.34}{3.4}

x = 5.1

8 0
3 years ago
Read 2 more answers
On a single set of axes, sketch a picture of the graphs of the following four equations: y = −x+ √ 2, y = −x− √ 2, y = x+ √ 2, a
Artist 52 [7]

Answer:

( 1/√ 2 , 1/√ 2 ) , ( 1/√ 2 , - 1/√ 2 ),  ( -1/√ 2 , 1/√ 2 ) , ( -1/√ 2 , - 1/√ 2 )  

y + 1 = - ( x + 2 ) + √ 2 , y + 1 = - ( x + 2 ) - √ 2 ,  y + 1 = ( x + 2 ) - √ 2

             y + 1 = ( x + 2 ) + √ 2  ,   ( x + 2 )^2 + ( y + 1)^2 = 1

Step-by-step explanation:

Given:

- Four functions to construct a diamond:

                y = −x+ √ 2,  y = −x− √ 2,  y = x+ √ 2, and y = x − √ 2.

Find:

a)Show that the unit circle sits inside this diamond tangentially; i.e. show that the unit circle intersects each of the four lines exactly once.

b)Find the intersection points between the unit circle and each of the four lines.

(c) Construct a diamond shaped region in which the circle of radius 1 centered at (−2, − 1) sits tangentially. Use the techniques of this section to help.

Solution:

- For first part see the attachment.

- The equation of the unit circle is given as follows:

                                      x^2 + y^2 = 1

- To determine points of intersection we have to solve each given function of y with unit circle equation for set of points of intersection:

                                For:  y = −x+ √ 2 , x - √ 2

                                And: x^2 + y^2 = 1

                                x^2 + (+/- * (x - √ 2))^2 = 1

                                x^2 + (x - √ 2)^2 = 1

                                2x^2 -2√ 2*x + 2 = 1

                                2x^2 -2√ 2*x + 1 = 0

                                 2[ x^2 - √ 2] + 1 = 0

Complete sqr:         (1 - 1/√ 2)^2 = 0

                                 x = 1/√ 2 , x = 1/√ 2                                          

                                 y = -1/√ 2 + √ 2 = 1/√ 2

                                 y = 1/√ 2 - √ 2 = - 1/√ 2

Points are:                ( 1/√ 2 , 1/√ 2 ) , ( 1/√ 2 , - 1/√ 2 )

- Using vertical symmetry of unit circle we can also evaluate other intersection points by intuition:

                                x = - 1/√ 2

                                 y = 1/√ 2 , -1/√ 2

Points are:              ( -1/√ 2 , 1/√ 2 ) , ( -1/√ 2 , - 1/√ 2 )  

- To determine the function for the rhombus region that would be tangential to unit circle with center at ( - 2 , - 1 ):

- To shift our unit circle from origin to ( - 2 , - 1 ) i.e two units left and 1 unit down.

- For shifts we use the following substitutions:

                           x = x + 2  ....... 2 units of left shift

                           y = y + 1 .......... 1 unit of down shift

- Now substitute the above shifting expression in all for functions we have:

                          y = −x+ √ 2 ----->  y + 1 = - ( x + 2 ) + √ 2

                          y = −x− √ 2 ----->  y + 1 = - ( x + 2 ) - √ 2

                          y = x- √ 2 ------->  y + 1 = ( x + 2 ) - √ 2

                          y = x+ √ 2 ------> y + 1 = ( x + 2 ) + √ 2

                          x^2 + y^2 = 1 ----->  ( x + 2 )^2 + ( y + 1)^2 = 1

- The following diamond shape graph would have the 4 functions as:

             y + 1 = - ( x + 2 ) + √ 2 , y + 1 = - ( x + 2 ) - √ 2 ,  y + 1 = ( x + 2 ) - √ 2

             y + 1 = ( x + 2 ) + √ 2  ,   ( x + 2 )^2 + ( y + 1)^2 = 1

- See attachment for the new sketch.            

7 0
3 years ago
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