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nika2105 [10]
3 years ago
11

D

Mathematics
2 answers:
aalyn [17]3 years ago
6 0

Answer: See image attached

Step-by-step explanation:

Gelneren [198K]3 years ago
3 0

Answer:

General equation of line : y = mx+c   --1

Where m is the slope or unit rate

Table 1)

p    d

1      3

2     6

4     12

d = Number of dollars (i.e.y axis)

p = number of pound(i.e. x axis)

First find the slope

First calculate the slope of given points

m = \frac{y_2-y_1}{x_2-x_1}   ---A

(x_1,y_1)=(1,3)

(x_2,y_2)=(2,6)

Substitute values in A

m = \frac{6-3}{2-1}

m = 3

Thus the unit rate is 3 dollars per pound.

So, It matches the box 1 (Refer the attached figure)

Equation 1 : p=3d

\frac{p}{3}=d

Since p is the x coordinate and d is the y coordinate

On Comparing with 1

m = \frac{1}{3}

Thus the unit rate is \frac{1}{3} dollars per pound

So, It matches the box 2 (Refer the attached figure)

Equation 2 : \frac{1}{3}d=3p

d=9p

Since p is the x coordinate and d is the y coordinate

On Comparing with 1

m =9

Thus the unit rate is 9 dollars per pound

So, It matches the box 3 (Refer the attached figure)

Table 2)

p        d

1/9      1

1         9

2        18

d = Number of dollars (i.e.y axis)

p = number of pound(i.e. x axis)

(x_1,y_1)=(\frac{1}{9},1)

(x_2,y_2)=(1,9)

Substitute values in A

m = \frac{9-1}{1-\frac{1}{9}}

m = \frac{8}{frac{8}{9}}

m = 9

Thus the unit rate is 9 dollars per pound

So, It matches the box 3 (Refer the attached figure)

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4. Gloria the grasshopper is working on her hops.
aivan3 [116]

The path that Gloria follows when she jumped is a path of parabola.

The equation of the parabola  that describes the path of her jump is \mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}

The given parameters are:

\mathbf{Height = 20}

\mathbf{Length = 28}

<em>Assume she starts from the origin (0,0)</em>

The midpoint would be:

\mathbf{Mid = \frac 12 \times Length}

\mathbf{Mid = \frac 12 \times 28}

\mathbf{Mid = 14}

So, the vertex of the parabola is:

\mathbf{Vertex = (Mid,Height)}

Express properly as:

\mathbf{(h,k) = (14,20)}

A point on the graph would be:

\mathbf{(x,y) = (28,0)}

The equation of a parabola is calculated using:

\mathbf{y = a(x - h)^2 + k}

Substitute \mathbf{(h,k) = (14,20)} in \mathbf{y = a(x - h)^2 + k}

\mathbf{y = a(x - 14)^2 + 20}

Substitute \mathbf{(x,y) = (28,0)} in \mathbf{y = a(x - 14)^2 + 20}

\mathbf{0 = a(28 - 14)^2 + 20}

\mathbf{0 = a(14)^2 + 20}

Collect like terms

\mathbf{a(14)^2 =- 20}

Solve for a

\mathbf{a =- \frac{20}{14^2}}

\mathbf{a =- \frac{20}{196}}

Simplify

\mathbf{a =- \frac{5}{49}}

Substitute \mathbf{a =- \frac{5}{49}} in \mathbf{y = a(x - 14)^2 + 20}

\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}

Hence, the equation of the parabola  that describes the path of her jump is \mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}

See attachment for the graph

Read more about equations of parabola at:

brainly.com/question/4074088

7 0
3 years ago
What is the measure of
Sonbull [250]

Answer:

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Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Use synthetic division and remainder theorem p(x) = 3x^3 - 5x^2 - x + 2. p(-1/3)=
Alik [6]

Answer:

5/3

Step-by-step explanation:

-1/3 goes on the outside and since we have our polynomial in standard form with no in between terms missing, 3,-5,-1,2 go inside because they are the coefficients of our polynomial.

-1/3   |   3      -5          -1          2

       |

         -------------------------------------

First step bring the 3 down inside. (3+0=3)

-1/3   |   3      -5          -1          2

       |

         -------------------------------------

           3

Whatever goes below the bar, must be multiplied by outside number and put directly below next number inside.

-1/3   |   3      -5          -1          2

       |             -1

         -------------------------------------

           3

The numbers lined up vertically are added to get the numbers underneath the bar.

-1/3   |   3      -5          -1          2

       |             -1

         -------------------------------------

           3        -6

Again any number below the bar gets multiply to the number outside.

-1/3   |   3      -5          -1          2

       |             -1          2

         -------------------------------------

           3        -6

Again the numbers lined up vertically above the bar get added to get the number that goes underneath the bar there.

-1/3   |   3      -5          -1          2

       |             -1          2

         -------------------------------------

           3        -6          1

Multiply to outside number 1(-1/3)=-1/3.

This goes under the 2 inside.

-1/3   |   3      -5          -1          2

       |             -1          2          -1/3

         -------------------------------------

           3        -6          1

The last number we are fixing to be put is the remainder of (3x^3-5x^2-x+2)/(x+1/3) or you could say it is the value of p(-1/3) since:

P(x)/(x-c)=Q(x)+R/(x-c)

Multiply both sides by (x-c):

P(x)=Q(x)(x-c)+R

If you evaluate P at x=c, we get R:

P(c)=Q(c)(c-c)+R

P(c)=Q(c)*0+R

P(c)=R.

Let's finish:

-1/3   |   3      -5          -1          2

       |             -1          2          -1/3

         -------------------------------------

           3        -6          1         5/3

This means p(-1/3)=5/3.

We could have also got this by directly plugging in (-1/3) for x into 3x^3-5x^2-x+2.

7 0
4 years ago
A quantity with an initial value of 980 decays exponentially at a rate of 0.3% every 6 decades. What is the value of the quantit
solong [7]

Answer:

956.73

Step-by-step explanation:

0.3% = 0.003

1-0.003 = 0.997

49/6 = 8 r1 (use 8 because decay is not continuous)

0.997^8 = 0.97625...

0.97625... x 980 ≈ 956.73

8 0
3 years ago
How would you prove this using a flow chart?
Genrish500 [490]

\triangle BCA\sim\triangle DEA\ (AA)\\\text{therefore the ratios of the lengths of their corresponding sides are equal}\\\\\dfrac{BC}{AC}=\dfrac{DE}{AE}\ \ \ |\text{cross multiply}\\\\BC\cdot AE=AC\cdot DE\ \ \ \ |:BC\\\\AE=\dfrac{AC\cdot DE}{BC}\ \ \ \ |:AC\\\\\dfrac{AE}{AC}=\dfrac{DE}{BC}\to\dfrac{DE}{BC}=\dfrac{AE}{AC}

6 0
4 years ago
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