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4 years ago

What is the solution to this inequality x-7<1

2 answers:

8 0

X−7<1

Add 7 to both sides.

x−7+7<1+7

x<8

To plot this on a number line, put a circle around 8 and a line going to the left with an arrow at the end.

Rashid [163]4 years ago

6 0

$x - 7 \ \textless \ 1$
$x \ \textless \ 7+1$
$x \ \textless \ 8$

Hope this helps !

Photon

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The question of two numbers is eight the sum of the same two numbers is 72 what are the two numbers

Otrada [13]

Answer:

The two numbers are 64 and 8.

Step-by-step explanation:

You meant "quotient" of two numbers, right?

Represent the numbers by x and y.

Then the quotient is x/y = 8, and beyond that we know that x + y = 72.

Solve this for x and y.

If x/y = 8, then x = 8y. Substituting this into x + y = 72, we get:

(8y) + y = 72, or 9y = 72. Dividing both sides by 9 yields y = 8.

Since x/y = 8, letting y = 8 results in x = 64.

The two numbers are 64 and 8.

5 0

3 years ago

Find the work required to move an object in the force field F = ex+y <1,1,z> along the straight line from A(0,0,0) to B(-1

storchak [24]

Answer:

Work = e+24

F is not conservative.

Step-by-step explanation:

To find the work required to move an object in the force field

$\large F(x,y,z)=(e^{x+y},e^{x+y},ze^{x+y})$

along the straight line from A(0,0,0) to B(-1,2,-5), we have to parameterize this segment.

Given two points P, Q in any euclidean space, you can always parameterize the segment of line that goes from P to Q with

r(t) = tQ + (1-t)P with 0 ≤ t ≤ 1

r(t) = t(-1,2,-5) + (1-t)(0,0,0) = (-t, 2t, -5t) with 0≤ t ≤ 1

is a parameterization of the segment.

the work W required to move an object in the force field F along the straight line from A to B is the line integral

$\large W=\int_{C}Fdr$

where C is the segment that goes from A to B.

$\large \int_{C}Fdr =\int_{0}^{1}F(r(t))\circ r'(t)dt=\int_{0}^{1}F(-t,2t,-5t)\circ (-1,2,-5)dt=\\\\=\int_{0}^{1}(e^t,e^t,-5te^t)\circ (-1,2,-5)dt=\int_{0}^{1}(-e^t+2e^t+25te^t)dt=\\\\\int_{0}^{1}e^tdt-25\int_{0}^{1}te^tdt=(e-1)+25\int_{0}^{1}te^tdt$

Integrating by parts the last integral:

$\large \int_{0}^{1}te^tdt=e-\int_{0}^{1}e^tdt=e-(e-1)=1$

and

$\large \boxed{W=\int_{C}Fdr=e+24}$

To show that F is not conservative, we could find another path D from A to B such that the work to move the particle from A to B along D is different to e+24

Now, let D be the path consisting on the segment that goes from A to (1,0,0) and then the segment from (1,0,0) to B.

The segment that goes from A to (1,0,0) can be parameterized as

r(t) = (t,0,0) with 0≤ t ≤ 1

so the work required to move the particle from A to (1,0,0) is

$\large \int_{0}^{1}(e^t,e^t,0)\circ (1,0,0)dt =\int_{0}^{1}e^tdt=e-1$

The segment that goes from (1,0,0) to B can be parameterized as

r(t) = (1-2t,2t,-5t) with 0≤ t ≤ 1

so the work required to move the particle from (1,0,0) to B is

$\large \int_{0}^{1}(e,e,-5et)\circ (-2,2,-5)dt =25e\int_{0}^{1}tdt=\frac{25e}{2}$

Hence, the work required to move the particle from A to B along D is

e - 1 + (25e)/2 = (27e)/2 -1

since this result differs from e+24, the force field F is not conservative.

6 0

4 years ago

What is the resulting equation when the expression for y in the second equation is substituted into the first equation? 3x y = 1

Natali5045456 [20]

After putting the value of y from the second equation to the first equation, the resultant equation is $x=5$ .

GIven:

The equations are:

$3x + y = 1\\
y = 6 - 4x$

It is required to put the value of y from second equation to the first equation.

<h3>How to solve equations?</h3>

The value of y from the second equation is,

$
y = 6 - 4x$

Now, put this value of y in the first equation as,

$3x + y = 1\\
3x+(6 - 4x)=1\\
3x+6-4x=1\\
-x=-5\\
x=5$

Therefore, after putting the value of y from the second equation to the first equation, the resultant equation is $x=5$ .

For more details about equations, refer to the link:

brainly.com/question/2263981

8 0

2 years ago

Please help me ASAP!

Whitepunk [10]

If they are similar, and the second cube's length is 3 times that of the first, then all of the sides are 3 times as large.
To calculate the volume, we multiply the length by the width by the height, each of which is 3 times as big. So, in total, the volume is 27 times larger in the second prism than the first.
So, if the first volume is 30 cubic centimetres, the second one would be 810 cubic centimetres.

Hope I helped! xx

4 0

3 years ago

Kisha bought a pair of shoes originally marked $48.35. The

anyanavicka [17]

Answer:

she saved $9.67

Step-by-step explanation:

$48.35÷ 0.20= $9.67

6 0

3 years ago