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2 years ago

Whats the solution for y=2x-4, y=3x-7

Mathematics

2 answers:

kodGreya [7K]2 years ago

8 0

The solution for y=2x-4: y= -8

The solution for y=3x-4: y= -12

Vikki [24]2 years ago

7 0

(Y,x) = (26,11)
Explained:

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How to find slope y=3x+5

klemol [59]

This equation is in slope-intercept form; y = mx+b.

m is the slope and b is the y-intercept, so to find the slope you'd look at the coefficient of x.

In this equation, y = 3x+5, the coefficient of x is 3, so that means that the slope of this equation is 3.

Answer is:

3 is the slope

3 0

3 years ago

a toy rocket is launched from the top of a 48 foot hill. The rockets initial upward velocity is 32 feet per second and its heigh

sdas [7]

Answer:

t = 2.11 seconds

Step-by-step explanation:

A toy rocket is launched from the top of a 48 foot hill. The rockets initial upward velocity is 32 feet per second and its height h at any given second t is modeled by the equation:

$h=-16t^2+32t+48$

Let us assume that we need to find the time by it to reach the ground. It means h = 0

$-16t^2+32t+48=0$

The above is a quadratic equation. The value of t is given by :

$t=\dfrac{-b\pm \sqrt{b^2-2ac} }{2a}\\\\t=\dfrac{-b+ \sqrt{b^2-2ac} }{2a},\dfrac{-b- \sqrt{b^2-2ac} }{2a}\\\\t=\dfrac{-32+ \sqrt{(32)^2-2\times (-16)(8)} }{2\times (-16)},\dfrac{-32-\sqrt{(32)^{2}-2\times(-16)(8)}}{2\times(-16)}\\\\t=-0.11\ s, 2.11\ s$

So, it will take 2.11 seconds to reach the ground.

6 0

3 years ago

Write a fourth grade word problem where the answer is 35.

rusak2 [61]

Answer:

Step-by-step explanation:

15+14+6

15+14= 29

29+6= 35

7 0

3 years ago

Read 2 more answers

Consider F and C below. F(x, y, z) = yz i + xz j + (xy + 6z) k C is the line segment from (2, 0, −2) to (5, 4, 2) (a) Find a fun

WITCHER [35]

Answer:

Required solution (a) $f(x,y,z)=xyz+3z^2+C$ (b) 40.

Step-by-step explanation:

Given,

$F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k$

(a) Let,

$F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k=f_x \uvec{i} +f_y \uvex{j}+f_z\uvec{k}$

Then,

$f_x=yz,f_y=xz,f_z=xy+6z$

Integrating $f_x$ we get,

$f(x,y,z)=xyz+g(y,z)$

Differentiate this with respect to y we get,

$f_y=xz+g'(y,z)$

compairinfg with $f_y=xz$ of the given function we get,

$g'(y,z)=0\implies g(y,z)=0+h(z)\implies g(y,z)=h(z)$

Then,

$f(x,y,z)=xyz+h(z)$

Again differentiate with respect to z we get,

$f_z=xy+h'(z)=xy+6z$

on compairing we get,

$h'(z)=6z\implies h(z)=3z^2+C$ (By integrating h'(z)) where C is integration constant. Hence,

$f(x,y,z)=xyz+3z^2+C$

(b) Next, to find the itegration,

$\int_C \vec{F}.dr=\int_C \nabla f. d\vec{r}=f(5,4,2)-f(2,0,-2)=(52+C)-(12+C)=40$

3 0

3 years ago

you have $895 in your checking account and write checks for $669, plus take $100 in cash out of your account. How much do you ha

Scilla [17]

To answer this we just need to set up a equation:-

895 - 669 - 100 = 126.

SO, after these transactions you will have $126 left.

Hope I helped ya!!

7 0

3 years ago