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horsena [70]
2 years ago
10

The diagram shows A(-3, 5) and B(8, 1) what is the distance between the points? Enter the answer as a number without units. Roun

d your answer to one decimal
Mathematics
1 answer:
lawyer [7]2 years ago
8 0

Answer:

AB=\sqrt{137}

Step-by-step explanation:

The Distance between two points in coordinate geometry can be find by the Distance formula:

If two points (x_1,y_1) and (x_2,y_2) are given:

Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

For points A(-3,5) and B(8,1)

        Distance (AB)=\sqrt{(8-(-3))^2+(1-5)^2}\\\\ Distance(AB)=\sqrt{(8+3)^2+(1-5)^2}\\\\Distance(AB)=\sqrt{11^2+(-4)^2}\\\\ Distance(AB)=\sqrt{121+16} \\\\AB=\sqrt{137}

The Distance of points is AB=\sqrt{137}

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alina1380 [7]

Answer:

Step-by-step explanation:

2 cups=8/4 1/2=2/4 1/8=0.5/4

8/4-1/4=7/4

7/4-2/4=5/4

5/4-0.5/4=4.5/4

Answer:1 0.5/4 cups

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2 years ago
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kow [346]

Answer:

For the first city, the 95% confidence interval would be:

28,900 +/- 2300 x 3 = 28,900 +/-6900$

For the second city, the 95% confidence interval would be:

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3 years ago
Select the correct answer.
Crazy boy [7]

check the picture below.

now, if we split that triangle like so with that red line, we can use the right-triangle there, to get θ.


\bf tan(\theta )=\cfrac{\stackrel{opposite}{3}}{\stackrel{adjacent}{3}}\implies tan(\theta )=1\implies tan^{-1}(\theta )=tan^{-1}(1)\implies \measuredangle \theta =\cfrac{\pi }{4}

8 0
2 years ago
Can you like solve dis fo me 40x(4x6^3+4x)
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160x^64+160x^2

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Step-by-step explanation:

6 0
3 years ago
The graph of an inverse trigonometric function passes through the point (1, pi/2). Which of the following could be the equation
rodikova [14]

Answer: C) y=sin^-1 x

Step-by-step explanation:

Since, the graph of an inverse trigonometric function will pass through the point (1,\frac{\pi}{2}),

If this point satisfies the function,

For the function y=cos^{-1} x

If x = 1

y=cos^{-1}1=0

Thus,  (1,\frac{\pi}{2}) is not satisfying function  y=cos^{-1} x,

⇒ The graph of   y=cos^{-1} x is not passing through the point  (1,\frac{\pi}{2})

For the function y=cot^{-1}x

If x = 1

y=cot^{-1}1=\frac{\pi}{4}

Thus,  (1,\frac{\pi}{2}) is not satisfying function y=cot^{-1}x,

⇒ The graph of   y=cot^{-1}x is not passing through the point  (1,\frac{\pi}{2})

For the function y=sin^{-1} x

If x = 1

y=sin^{-1}1=\frac{\pi}{2}

Thus,  (1,\frac{\pi}{2}) is satisfying function y=sin^{-1} x,

⇒ The graph of   y=sin^{-1} x is passing through the point  (1,\frac{\pi}{2}).

For the function y=tan^{-1}x

If x = 1

y=tan^{-1}1=\frac{\pi}{4}

Thus,  (1,\frac{\pi}{2}) is not satisfying function  y=cos^{-1} x,

⇒ The graph of   y=tan^{-1} x is not passing through the point (1,\frac{\pi}{2}).

Hence, Option C is correct.

3 0
3 years ago
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