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Mars2501 [29]
2 years ago
7

Question 2 11 + 5x = 5x + 20

Mathematics
1 answer:
julia-pushkina [17]2 years ago
7 0

Answer:

11+5x=5x+20

We move all terms to the left:

11+5x-(5x+20)=0

We get rid of parentheses

5x-5x-20+11=0

We add all the numbers together, and all the variables

-9!=0

There is no solution for this equation

Step-by-step explanation:

I saw it from Google

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Answer:

13

Explanation:
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3 years ago
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what are the lengths of the legs of a right triangle in which one acute angle measures 19° and the hypotensue is 15 units long?
givi [52]
Refer to the diagram shown below.

Given:
m∠A = 19°
c = 15

By definition,
sin A = a/c
Therefore
a = c*sin A = 15*sin(19°) = 4.8835

cos A = b/c
Therefore
b = c*cos A = 15*cos(19°) =14.1828

Answer:
The lengths are 4.88, 14.18, and 15.00  (nearest hundredth)  

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3 years ago
Find a cubic function with the given zeros.
Fed [463]

Answer:

The correct option is D) f(x) = x^3 + 2x^2 - 2x - 4 .

Step-by-step explanation:

Consider the provided cubic function.

We need to find the equation having zeros: Square root of two, negative Square root of two, and -2.

A "zero" of a given function is an input value that produces an output of 0.

Substitute the value of zeros in the provided options to check.

Substitute x=-2 in f(x) = x^3 + 2x^2 - 2x + 4 .

f(x) = x^3 + 2x^2 - 2x + 4\\f(x) = (-2)^3 + 2(-2)^2 - 2(-2) + 4\\f(x) =-8 + 2(4)+4 + 4\\f(x) =8

Therefore, the option is incorrect.

Substitute x=-2 in f(x) = x^3 + 2x^2 + 2x - 4 .

f(x) = x^3 + 2x^2 + 2x - 4\\f(x) = (-2)^3 + 2(-2)^2 + 2(-2) - 4\\f(x) =-8+2(4)-4-4\\f(x) =-8

Therefore, the option is incorrect.

Substitute x=-2 in f(x) = x^3 - 2x^2 - 2x - 4 .

f(x) = x^3 - 2x^2 - 2x - 4\\f(x) = (-2)^3 - 2(-2)^2 - 2(-2) - 4\\f(x) =-8-8+4-4\\f(x) =-16

Therefore, the option is incorrect.

Substitute x=-2 in f(x) = x^3 + 2x^2 - 2x - 4 .

f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (-2)^3+2(-2)^2 - 2(-2) - 4\\f(x) =-8+8+4-4\\f(x) =0

Now check for other roots as well.

Substitute x=√2 in f(x) = x^3 + 2x^2 - 2x - 4 .

f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (\sqrt{2})^3+2(\sqrt{2})^2 - 2(\sqrt{2}) - 4\\f(x) =2\sqrt{2}+4-2\sqrt{2}-4\\f(x) =0

Substitute x=-√2 in f(x) = x^3 + 2x^2 - 2x - 4 .

f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (-\sqrt{2})^3+2(-\sqrt{2})^2 - 2(-\sqrt{2}) - 4\\f(x) =-2\sqrt{2}+4+2\sqrt{2}-4\\f(x) =0

Therefore, the option is correct.

8 0
3 years ago
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Hope this helps ^^

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The difference between vertical angles and linear pairs of angles is that linear pairs of angles has two adjacent angles whose noncommon sides form opposite sides and vertical angles have two adjacent angles that have common sides that and they have a common vertex.

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3 years ago
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