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3 years ago

Write the given sentence as an equation. Tim’s age in 7 years will be three times what it was 19 years ago.

1 answer:

5 0

T +7 is Tim's age in 7 years
three times what it was 19 years ago = 3(t-19)

so equation should be:
t + 7 = 3(t - 19)

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100 Points and Brainly.

user100 [1]

function : y = (-x) - 6

Find x-intercept :

(-x) - 6 = 0

(-x) = 6

x = -6

Find y-intercept :

y = (-0) - 6

y = -6

mark these two points on both the axis and draw a straight linear graph.

passes coordinates : (0, -6), (-6, 0)

6 0

2 years ago

Read 2 more answers

What is the slope of the line that passes through the points (-20, 18) and (30, 14)? 25

vodomira [7]

Answer:

The equation of the line that passes through the points

(-20,18) and (30,14)

y=-2/25x+82/5

6 0

2 years ago

find two numbers that round to 15.5 when rounded to the nearest tenth. besides 15.04 15.55 15.508 15.445 15.0 15.49

Kitty [74]

Two other numbers are 15.48 and 15.52

4 0

2 years ago

Graph the line whose y-intercept is -7 and whose x-intercept is 1

vesna_86 [32]

Answer:

$y=7x -7$

Step-by-step explanation:

We are given two points, the y - intercept and the x - intercept.

The y - intercept being: ( 0, -7 )

The x - intercept being: ( 1, 0 )

The equation of a line in slope - intercept form is:

y = mx + b, where m is the slope and b is the y - intercept.

We already have the y - intercept. So far we have:

y = mx - 7

What we're missing is the slope.

To find the slope, we can use this formula:

$\frac{y_{2}-y_{1} }{x_{2}-x_{1}}$ , where $x_{2}$ and $x_{1}$ are the x - coordinates of both points and $y_{2}$ and $x_{1}$ are the y - coordinates of both points.

$m=\frac{0-(-7)}{1-0} \\\\m=\frac{7}{1}\\m=7$

We now have the missing piece, the slope, so the equation now is:

$y=7x -7$

3 0

2 years ago

A potential energy function is given by U = A x3 + B x2 − C x + D . For positive parameters A, B, C, D this potential has two eq

Ivenika [448]

Answer:

7.8655

Step-by-step explanation:

Given the functionl

$U=Ax^3+Bx^2-cx+D$

Where,

$A=1.45J/m^3\\B=2.85J/m^2\\c= 1.7J/m\\D=0.6J$

We have, replacing,

$U=1.45x^3+2.85x^2-1.7x+0.6$

We need to derivate and verify for stable equilibrum that,

$\frac{d^2 U}{dx^2}>0$

Then,

$U'(x) = 4.32x^2+5.7x-1.7$

Roots in,

$x_1=-1.57008\\x_2=0.250636$

We obtain U"(x), then we have

$\frac{d^2U}{dx}=8.64x+5.7$

For $x=-1.57, U"(x)=-7.86549$ Unestable

For $x=0.250636 U"(x)=7.8655>0$ Stable

7 0

2 years ago