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Zielflug [23.3K]

3 years ago

15

taka is playing a game of darts. she scores 52 points. in the game she is playing, she gets a -3 score each time she misses the

board. her final point total is 37. write and solve an equation to find out how many times she missed the board

1 answer:

Anon25 [30]3 years ago

7 0

Answer:

5

Step-by-step explanation:

52-37 is 15 . divide by 3 and u get five

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Solvex + 5-6 = 7.

yKpoI14uk [10]

I got x=8
Because 5-6=-1….. 7- -1= 8…… then it’s just that.I

I don’t understand the answers given tho but that’s my guess

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Answer:

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Step-by-step explanation:

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3 years ago

What is 1/6+-2/3-5/12=

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4 years ago

Solve the following system of equations using the substitution method. y = –5x – 17 –3x – 3y = 3

jek_recluse [69]

The answer to the question

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3 years ago

Special right triangles

garik1379 [7]

Answer: The answers are given below.

Step-by-step explanation: The calculations are as follows.

(1) We have in the given right-angled triangle,

$\tan 30^\circ=\dfrac{10}{x}\\\\\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{10}{x}\\\\\Rightarrow x=10\sqrt3,$

and

$y^2=(10)^2+x^2=100+(10\sqrt3)^2=100+300=400\\\\\Rightarrow y=20.$

∴ x = 10√3 and y = 20.

(2) We have in the given right-angled triangle,

$\cos 60^\circ=\dfrac{2}{x}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{2}{x}\\\\\Rightarrow x=4,$

and

$y^2=x^2-2^2=16-4=12\\\\\Rightarrow y=2\sqrt3.$

∴ x = 4 and y = 2√3.

(3) We have in the given right-angled triangle,

$\cos 30^\circ=\dfrac{7}{y}\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{7}{y}\\\\\Rightarrow y=\dfrac{14\sqrt3}{3},$

and

$\sin 30^\circ=\dfrac{7}{x}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{7}{x}\\\\\Rightarrow x=14.$

∴ x = 14 and y = 14√3.

(4) We have in the given right-angled triangle,

$\sin 30^\circ=\dfrac{x}{6}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{x}{6}\\\\\Rightarrow x=3,$

and

$\cos 30^\circ=\dfrac{y}{6}\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{y}{6}\\\\\Rightarrow y=3\sqrt3.$

∴ x = 3 and y = 3√3.

(5) We have in the given right-angled triangle,

$\sin 30^\circ=\dfrac{y}{10}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{y}{10}\\\\\Rightarrow y=5,$

and

$\cos 30^\circ=\dfrac{x}{6}\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{x}{10}\\\\\Rightarrow y=5\sqrt3.$

∴ x = 5 and y = 5√3.

(6) We have in the given right-angled triangle,

$\sin 30^\circ=\dfrac{x}{8}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{x}{8}\\\\\Rightarrow x=4,$

and

$\cos 30^\circ=\dfrac{y}{8}\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{y}{8}\\\\\Rightarrow y=4\sqrt3.$

∴ x = 4 and y = 4√3.

(7) We have in the given right-angled triangle,

$\cos 30^\circ=\dfrac{7\sqrt3}{y}\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{7\sqrt3}{y}\\\\\Rightarrow y=14,$

and

$\sin 30^\circ=\dfrac{x}{y}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{x}{14}\\\\\Rightarrow x=7.$

∴ x = 7 and y = 14.

(8) We have in the given right-angled triangle,

$\cos 30^\circ=\dfrac{6\sqrt3}{y}\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{6\sqrt3}{y}\\\\\Rightarrow y=12$

and

$\sin 30^\circ=\dfrac{x}{y}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{x}{12}\\\\\Rightarrow x=6$

∴ x = 6 and y = 12.

(9) We have in the given right-angled triangle,

$\cos 30^\circ=\dfrac{\sqrt3}{y}\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{\sqrt3}{y}\\\\\Rightarrow y=2$

and

$\sin 30^\circ=\dfrac{x}{y}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{x}{2}\\\\\Rightarrow x=4.$

∴ x = 4 and y = 2.

Thus, all are completed.

3 0

3 years ago