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3 years ago

What is the sum of 6 and w is greater than or equal to 17 as an inequality

Mathematics

1 answer:

MaRussiya [10]3 years ago

7 0

Answer:

$11 ≤ w$

Step-by-step explanation:

w + 6 ≥ 17

- 6 - 6

_______

$w ≥ 11$

** The above answer is written in reverse, which is the exact same result.

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(3x2 – 5x + 6) + (2x2 + 7x - 9)

tatuchka [14]

Answer:

Step-by-step explanation:

3x^2 + 2x^2 - 5x + 7x + 6 - 9

5x^2 + 2x - 3

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3 years ago

Determine whether the relation is or is not a function.

soldier1979 [14.2K]

Answer:

Not a function because it does not pass the Vertical Line Test.

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3 years ago

If x is a positive integer, which expression is equivalent to StartFraction RootIndex 4 StartRoot x cubed EndRoot Over RootIndex

il63 [147K]

Answer:

^4 sqrt x^3/ ^5 sqrt x^2

Step-by-step explanation:

I got that on my test.

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3 years ago

Read 2 more answers

It is known that the straight line l is tangent to the circle x^2+y^2=4 at a point on the x-axis, and intersects with the straig

Talja [164]

Step-by-step explanation:

The general equation of a circle is

$(x \ - \ h)^2 \ + \ (y \ - \ k)^2 \ = \ r^{2}$ ,

where h and k forms the coordinates of the centre of the circle.

When the circle has a centre at the origin, the equation reduces into

. $x^2 \ + \ y^2 \ = r^2$ .

Now, we are interested in solving for the x-intercepts (the x-coordinates when the circle intersects the x-axis), of the circle

$x^2 \ + \ y^2 \ = \ 4$ .

Thus,

$x^2 \ + \ (0)^2 \ = \ 4 \\ \\ \\ \-\hspace{1.3cm} x^{2} \ = \ 4 \\ \\ \\ \-\hspace{1.4cm} x \ = \ \pm \ 2$ .

Geometrically speaking, the tangent to the circle at the point defined by one of the x-intercepts of the circle is actually a vertical line, more specifically the lines $x \ = \ \pm \ 2$ .

First and foremost, for the vertical line $x \ = \ 2$ , it intersects the straight line $y \ = \ \displaystyle\frac{1}{2}x$ , giving the y-coordinate for point P,

$y \ = \ \displaystyle\frac{1}{2}(2) \\ \\ \\ y \ = \ 1$ .

Hence, the coordinates of point P are $(1, \ 1)$ .

However, since there are no boundaries given in the question and a circle is symmetrical about its centre, thus, point P also exists when the vertical line $x \ = \ -2$ and interdects the straight line $y \ = \ \displaystyle\frac{1}{2}x$ .

$y \ = \ \displaystyle\frac{1}{2}(-2) \\ \\ \\ y \ = \ -1$ .

Therefore, the coordinates of point P are also $(1, \ -1)$ .

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2 years ago

3-10- -6+5+-7 whats the answer

shusha [124]

1. Simplify brackets (3 - 10 + 6 + 5 - 7)

2. Simplify.

Answer: -3.

7 0

3 years ago