All categories

1 year ago

What is the value of x?

Mathematics

2 answers:

nalin [4]1 year ago

7 0

ANSWER: x = 27

EXPLANATION: 2 angles opposite from each other at the vortex are equal. You would set these equations equal to each other.

1. set them equal to each other.
1. (4x+7) = [5(x+l-4)].
2. Distribute the 5
2. (4x+7) = (5x-20)
3. isolate the variable (add 20 to 7)
3. 4x + 27 = 5x
4. combine the x’s (subtract 4x from 5x)
4. 27 = x

Now, to be 100% sure of your answer, fill in the x’s.

4(27) + 7 = 5[(27)-4]
108 + 7 = 5[23]
115 = 115

ki77a [65]1 year ago

3 0

X = 27
cant explain since i looked it up but im sure you can buddy

You might be interested in

One cell phone company offers a plan that costs $26.99 and includes unlimited night and weekend minutes. Another company offers

Nutka1998 [239]

Given :

One cell phone company offers a plan that costs $26.99 and includes unlimited night and weekend minutes.

Another company offers a plan that costs $18.99 and charges $0.35 per minute during nights and weekends.

To Find :

For what numbers of night and weekend minutes does the second company's plan cost more than the first company's plan.

Solution :

Let, after n number of nights and weekends second plan is most costly.

So,

$18.99+0.35x > 26.99\\\\0.35x > 8\\\\x>22.85$

Therefore, after 22 night and weekends second plan is most costly.

Hence, this is the required solution.

8 0

3 years ago

Go step by step to reduce the radical of the equation given. (Figure the step for the box)

Rama09 [41]

$\sqrt{32} ~~ \begin{cases} 32=2\cdot 2\cdot 2\cdot 2\cdot 2\\ \qquad 2^4\cdot 2\\ \qquad 2^{(2)(2)}\cdot 2\\ \qquad (2^2)^2\cdot 2 \end{cases}\implies \sqrt{(2^2)^2\cdot 2}\implies 2^2\sqrt{2}\implies 4\sqrt{2}$

7 0

2 years ago

It is a best practice to analyze data directly on an infected machine. t or f?

Alex Ar [27]

I would say your answer is FALSE

5 0

3 years ago

The scatter plot below shows the relationship between the diameter (x), in mm, and mass f(x), in grams, of a berry: ordered pair

kkurt [141]

The function would be f (x) = 0.4x - 4.

Explanation
We can see that this is linear, since it increases at a constant rate. We find the slope using the formula
m=(y₂-y₁)/(x₂-x₁) = (0.8-0.4)/(12-11) = 0.4/1 = 0.4

We use point-slope form to write the initial equation:

y-y₁ = m(x-x₁)
y-0.4 = 0.4(x-11)

Using the distributive property:
y-0.4 = 0.4*x - 0.4*11
y-0.4 = 0.4x-4.4

Add 0.4 to both sides:
y-0.4+0.4 = 0.4x-4.4+0.4
y=0.4x-4

8 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