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juin [17]
3 years ago
7

Which expression is equalvalant to 6x+8?

Mathematics
2 answers:
Alenkasestr [34]3 years ago
8 0
In order to solve this you must factor it out. After factoring you come to the answer-
B) 2(3x+4)
Westkost [7]3 years ago
4 0
The answer is B

2 (3x+4)
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What is the equation of a line, in general form, that passes through point (1, -2) and has a slope of 1/3. 1.x - 3 y - 7 = 2.x -
Nutka1998 [239]

Answer:

x- 3y - 7=0

Step-by-step explanation:

the line in the slope-intercept form:

  • y= mx+b

m= 1/3 and point given (1, -2) by using which we find y-intercept:

  • b= y-1/3x
  • b= -2- 1/3*1= - 7/3
  • y= 1/3x- 7/3

General form of a line is: ax+by+c= 0

converting:

  • y= 1/3x- 7/3  ⇒  3y= x- 7  ⇒  x- 3y - 7=0
5 0
3 years ago
I need help on this?!
Sergio039 [100]
If you nee solving it will be:
c^6(-3c^5)^2
c^6(3c^5)^2
c^6·3^2(c^5)^2
c^6·9(c^5)^2
c^6·9c^10
9c^6c^10
9c^6+10
9c^16
So the answer is:
9c^16
(9c to the power of 16)
Sorry I was late..
3 0
3 years ago
Denzel has 3 2/5 boxes of party favors. One full box contains 15 bags of favors and each bag has 3 favors in it. Which expressio
Artist 52 [7]

Boxes = 3_2/5

15 = total number of bags of favors

Each bag has 3 items inside

3 × 15 = 45

Total number of items = number of boxes plus (15 × 3).

So, choice A is the answer.

5 0
3 years ago
Verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given poi
lara [203]

For each curve, plug in the given point (x,y) and check if the equality holds. For example:

(I) (2, 3) does lie on x^2+xy-y^2=1 since 2^2 + 2*3 - 3^2 = 4 + 6 - 9 = 1.

For part (a), compute the derivative \frac{\mathrm dy}{\mathrm dx}, and evaluate it for the given point (x,y). This is the slope of the tangent line at the point. For example:

(I) The derivative is

x^2+xy-y^2=1\overset{\frac{\mathrm d}{\mathrm dx}}{\implies}2x+x\dfrac{\mathrm dy}{\mathrm dx}+y-2y\dfrac{\mathrm dy}{\mathrm dx}=0\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x+y}{2y-x}

so the slope of the tangent at (2, 3) is

\dfrac{\mathrm dy}{\mathrm dx}(2,3)=\dfrac74

and its equation is then

y-3=\dfrac74(x-2)\implies y=\dfrac74x-\dfrac12

For part (b), recall that normal lines are perpendicular to tangent lines, so their slopes are negative reciprocals of the slopes of the tangents, -\frac1{\frac{\mathrm dy}{\mathrm dx}}. For example:

(I) The tangent has slope 7/4, so the normal has slope -4/7. Then the normal line has equation

y-3=-\dfrac47(x-2)\implies y=-\dfrac47x+\dfrac{29}7

3 0
3 years ago
SOOOOO WHATCHA THINK??
Harman [31]

Answer:

x ≥20

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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