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3 years ago

Answer this questions right and i will give you Brainliest, Thanks, and 5 stars and thanks on profile.

Mathematics

1 answer:

katen-ka-za [31]3 years ago

3 0

Answer:

1. k=5

2. a= -7

3. n=4

4. n=1

5.x=0

6. r= -4

7. b=31

8.n= -3

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at the local baseball diamond,the distance from home base to second base is 100 feet. Determine the distance from home base to f

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Answer:

70.7 ft

Step-by-step explanation:

The triangle formed by home plate, first base, and second base is an isosceles right triangle with hypotenuse 100 ft. Then the side length (base-line distance between bases) is (100 ft)/√2 ≈ 70.7 ft.

___

For an isosceles right triangle with side lengths 1, the hypotenuse can be found from the Pythagorean theorem to be ...

hypotenuse = √(side² + side²) = √(1²+1²) = √2

Then the diamond distances satisfy the proportion ...

hypotenuse/side = 100 ft/(base distance) = √2/1

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base distance = (100 ft)/√2 ≈ 70.71068 ft

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If the coefficient of determination for a data set is 0.75 and the SSE for the data set is 9, what is the SST for the data set?

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Quadrilateral ABCD has the following vertices: A(-4, -3), B(2, -3), C(4, -6), and D(-4, -6). Quadrilateral ABCD is a

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In order to get the figure, you have to plot it.

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Find the exact value of each trigonometric function for the given angle θ.

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Answer:

$\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=\dfrac{2}{\sqrt{3}}.$

Step-by-step explanation:

The given angle is 240 degrees.

We need to find the exact value of each trigonometric function for the given angle θ.

Since $\theta=240$ , it means θ lies in 3rd quadrant. In 3d quadrant only tan and cot are positive.

$\sin (240^\circ)=\sin (180^\circ+60^\circ)=-\sin (60^\circ)=-\dfrac{\sqrt{3}}{2}$

$\cos (240^\circ)=\cos (180^\circ+60^\circ)=-\cos (60^\circ)=-\dfrac{1}{2}$

$\tan (240^\circ)=\tan (180^\circ+60^\circ)=\tan (60^\circ)=\sqrt{3}$

$\cot (240^\circ)=\cot (180^\circ+60^\circ)=\cot (60^\circ)=\dfrac{1}{\sqrt{3}}$

$\sec (240^\circ)=\sec (180^\circ+60^\circ)=-\sec (60^\circ)=-2$

$\csc (240^\circ)=\csc (180^\circ+60^\circ)=-\csc (60^\circ)=-\dfrac{2}{\sqrt{3}}$

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