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kifflom [539]
3 years ago
14

Match the equation with property used

Mathematics
1 answer:
nata0808 [166]3 years ago
7 0
I dont really know the difference between them lol; also it's a little blurry



sorry
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Someone help me asap please.
zaharov [31]

15. x ≤ -24

    x + 19 ≤ -5

    x ≤ -24

16. z > -12

    3z > -36

    z > -12

17. 185 + c > 410; c > 225


18.   C.

    x - 6 ≥ 1

    x ≥ 7

7 0
3 years ago
Brenda throws a dart at this square-shaped target: Part A: Is the probability of hitting the black circle inside the target clos
Masteriza [31]
Either way.  The probability of hitting the circle is:

P(C)=Area of circle divided by area of square

P(W)=(area of square minus area of circle divided by area of square

P(C)=(πr^2)/s^2

P(W)=(s^2-πr^2)/s^2

...

Okay with know dimensions, r=1 (because r=d/2 and d=2 so r=1), s=11 we have:

P(inside circle)=π/121  (≈0.0259  or 2.6%)

P(outside circel)=(121-π)/121  (≈0.9744 or 97.4%)
6 0
3 years ago
35 divided by 4 using remainders
tester [92]

Answer:

8 Remainder 2

Step-by-step explanation:


6 0
3 years ago
PLEASE HELP [PLEASE] YOUR SO NICE. BRAINLIEST WILL BE REWARDED PLASE HELP
s344n2d4d5 [400]
The correct answer is D)
5 0
3 years ago
Where should the point P be chosen on line segment AB so as to maximize the angle θ? (Assume a = 4 units, b = 5 units, and c = 9
taurus [48]
From the figure, let the distance of point P from point A on line segment AB be x and let the angle opposite side a be M and the angle opposite side c be N.

Using pythagoras theorem,
\tan M= \frac{a}{b-x} \\ \\ M=\tan^{-1}\left(\frac{a}{b-x}\right)
and
\tan N= \frac{c}{x} \\ \\ N=\tan^{-1}\left(\frac{c}{x}\right)

Angle θ is given by
\theta=180-M-N \\  \\ =180-\tan^{-1}\left(\frac{a}{b-x}\right)-\tan^{-1}\left(\frac{c}{x}\right)

Given that a = 4 units, b = 5 units, and c = 9 units, thus
\theta=180-\tan^{-1}\left(\frac{4}{5-x}\right)-\tan^{-1}\left(\frac{9}{x}\right)

To maximixe angle θ, the differentiation of <span>θ with respect to x must be equal to zero.
i.e.
\frac{d\theta}{dx} = -\frac{4}{x^2-10x+41} + \frac{9}{x^2+81} =0 \\  \\ -4(x^2+81)+9(x^2-10x+41)=0 \\  \\ -4x^2-324+9x^2-90x+369=0 \\  \\ 5x^2-90x+45=0 \\  \\ x^2-18x+9=0 \\  \\ x=9\pm6 \sqrt{2}

Given that x is a point on line segment AB, this means that x is a positive number less than 5.

Thus
x=9-6 \sqrt{2}=0.5147

Therefore, The distance from A of point P, so that </span>angle θ is maximum is 0.51 to two decimal places.
6 0
3 years ago
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