All categories

4 months ago

What is the equation of a line that passes through (4, 3) and has a slope of 2?

Mathematics

1 answer:

Savatey [412]4 months ago

4 0

Passes through (4,3) slope = 2

y - y1 = m(x - x1)
y - 3 = 2(x - 4)
y - 3 = 2x - 8
y = 2x - 5

C

You might be interested in

(82 POINTS AND BRAINLYEST)

algol13

Answer:

4 inches

Step-by-step explanation:

The equation for volume of a rectangular prism is V=lwh, where l=length, w=width, h=height.

We are given the volume, length, and width, so let's put those in and solve for the height.

V = lwh

660.24 = 13.1*12.6*h

660.24=165.06h

660.24/165.06 = h

4 = h

<u>The height is 4 inches.</u>

7 0

3 years ago

Read 2 more answers

Help! Please show work, and identify the solution or how many solutions there are to the system.

Sveta_85 [38]

Answer:

one solution (-3, -6)

Step-by-step explanation:

the two line cross at the point (-3, -6) and only at that one point so there is only one solution (-3, -6).

4 0

2 years ago

Please help will give brainliest!! thank u

Goshia [24]

9514 1404 393

Answer:

Step-by-step explanation:

The centroid divides a median into parts with the ratio 1:2, so RL:LD = 1:2. Then RL:RD = 1:(1+2), and RD = 3RL.

RD = 3·54 cm

RD = 162 cm

5 0

2 years ago

Find the product. (4m^2 – 5)(4m^2 + 5)

aliya0001 [1]

Answer: 16m^4−25

Step-by-step explanation:

(4m^2−5)(4m^2+5)

=(4m^2+−5)(4m^2+5)

=(4m^2)(4m^2)+(4m^2)(5)+(−5)(4m^2)+(−5)(5)

=16m^4+20m^2−20m^2−25

=16m^4−25

3 0

2 years ago

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

2 years ago