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zvonat [6]
3 years ago
12

ILL GIVE BRAINLEST

Mathematics
2 answers:
Anika [276]3 years ago
8 0

Answer:

Not a function

Step-by-step explanation:

(0,3,4,4,5)

lukranit [14]3 years ago
4 0

Answer:

not a function

Step-by-step explanation:

3 - 2 = 1

2 - 5 = -3

5-6 = -1

6-7 = -1

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What is the solution to the system of equations that contains −3x + y = 3 and 2x − y = −1? no solution (−1, 3) infinite number o
Paraphin [41]

Answer:

A) −3x + y = 3

B) 2x − y = −1  Adding both equations:

-x = 2

x = -2  y = -3

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
~PLEASE HELP ASAP OFFERING 12 POINTS~
Anna35 [415]

The answer is definitely 60%

7 0
3 years ago
Solve for x: 4 over 5 x + 4 over 3 = 2x x =
Naily [24]

Answer:

over 5 x + 4 over 3 = 2x

can't find x

Hi McDa:

Here's a couple hints. (I have no idea what you already understand. Please check out our posting guidelines.)

Express the Natural number 3 in Rational form 3/1.

4

5

x

+

4

3

1

=

2

x

Regarding the left-hand side above, have you learned the property that says we may multiply by the reciprocal of 3/1, instead of dividing by 3/1?

Step-by-step explanation: Incorrect? Let me know!

8 0
3 years ago
Let O be an angle in quadrant III such that cos 0 = -2/5 Find the exact values of csco and tan 0.​
vivado [14]

well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus

cos(\theta)=\cfrac{\stackrel{adjacent}{-2}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2 - (-2)^2}=b\implies \pm\sqrt{25-4}\implies \pm\sqrt{21}=b\implies \stackrel{III~Quadrant}{-\sqrt{21}=b}

\dotfill\\\\ csc(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{-\sqrt{21}}}\implies \stackrel{\textit{rationalizing the denominator}}{-\cfrac{5}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies -\cfrac{5\sqrt{21}}{21}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-\sqrt{21}}}{\underset{adjacent}{-2}}\implies tan(\theta)=\cfrac{\sqrt{21}}{2}

4 0
2 years ago
At a food festival, 3/8 of the dishes were from china. Another 12.5% of the dishes were from japan. What percent of the dishes w
MrMuchimi

Answer: 50%

Step-by-step explanation:

Let x = Total dishes.

Dishes from China = \dfrac38x

Dishes from Japan  = 12.5\% \text{ of }x= 0.125x

Total dishes  from China and Japan = \dfrac38 x+0.125x=0.375x+0.125x=0.5x

Dishes from other countries = x- 0.5x= 0.5x

Percent of dishes from other countries= \dfrac{0.5x}{x}\times100\%= 50\%

Hence, dishes from other countries = 50%

5 0
3 years ago
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