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olchik [2.2K]
2 years ago
10

For which function is the ordered pair (5, 10) not a solution?

Mathematics
1 answer:
Vitek1552 [10]2 years ago
7 0

Answer:

y = x - 5

Step-by-step explanation:

y = x - 5

10 = 5 - 5

10 is not equal 0

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What are the solutions to the equation -5x2 + 3x = -9
Aleks04 [339]

Answer: Move terms to the left side−52+3=−9−5x2+3x=−9−52+3−(−9)=0−


Common factor−52+3+9=0−5x2+3x+9=0−(52−3−9)=0


Divide both sides by the same factor−(52−3−9)=0−(5x2−3x−9)=052−3−9=0


Solution=3±321 over 10

Step-by-step explanation:

7 0
2 years ago
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Can someone help me with this?​
Anna35 [415]

Answer:

Let 'a' be the first term, 'r' be the common ratio and 'n' be the number of terms

Series = 2+6+18.......= 2+2•3¹+ 2•3².......= 728

Now,

Sum =  \frac{a( {r}^{n} - 1) }{(r - 1)}   \\

So,

\frac{a( {r}^{n} - 1)}{(r - 1)}  = 728 \\  \frac{2( {3}^{n} - 1) }{(3 - 1)}  = 728 \\  \frac{2( {3}^{n} - 1) }{2}  = 728 \\  {3}^{n}  - 1 = 728 \\ {3}^{n}=728+1\\ {3}^{n}  = 729 \\  {3}^{n}  =  {3}^{6}  \\ \boxed{ n = 6}

Therefore, number of terms is 6

  • 6 is the right answer.
3 0
2 years ago
Norm has $15,000 to deposit. His daughter is a junior in high school and plans to go to college. Recommend the best way for Norm
Aloiza [94]
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8 0
3 years ago
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i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.
Natali [406]

Answer:

sin\ \varnothing = \frac{-1}{10}\sqrt 2

cos\ \varnothing = \frac{-7}{10}\sqrt 2

tan\ \varnothing = \frac{1}{7}

cot\ \varnothing = 7

sec\ \varnothing = \frac{-5}{7}\sqrt 2

csc\ \varnothing = -5\sqrt 2

Step-by-step explanation:

Given

N = (-7,-1) --- terminal side of \varnothing

Required

Determine the values of trigonometric functions of \varnothing.

For \varnothing, the trigonometry ratios are:

sin\ \varnothing = \frac{y}{r}       cos\ \varnothing = \frac{x}{r}       tan\ \varnothing = \frac{y}{x}

cot\ \varnothing = \frac{x}{y}       sec\ \varnothing = \frac{r}{x}       csc\ \varnothing = \frac{r}{y}

Where:

r^2 = x^2 + y^2

r = \sqrt{x^2 + y^2

In N = (-7,-1)

x = -7 and y = -1

So:

r = \sqrt{(-7)^2 + (-1)^2

r = \sqrt{50

r = \sqrt{25 * 2

r = \sqrt{25} * \sqrt 2

r = 5 * \sqrt 2

r = 5 \sqrt 2

<u>Solving the trigonometry functions</u>

sin\ \varnothing = \frac{y}{r}

sin\ \varnothing = \frac{-1}{5\sqrt 2}

Rationalize:

sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}

sin\ \varnothing = \frac{-\sqrt 2}{5*2}

sin\ \varnothing = \frac{-\sqrt 2}{10}

sin\ \varnothing = \frac{-1}{10}\sqrt 2

cos\ \varnothing = \frac{x}{r}

cos\ \varnothing = \frac{-7}{5\sqrt 2}

Rationalize

cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}

cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}

cos\ \varnothing = \frac{-7\sqrt 2}{10}

cos\ \varnothing = \frac{-7}{10}\sqrt 2

tan\ \varnothing = \frac{y}{x}

tan\ \varnothing = \frac{-1}{-7}

tan\ \varnothing = \frac{1}{7}

cot\ \varnothing = \frac{x}{y}

cot\ \varnothing = \frac{-7}{-1}

cot\ \varnothing = 7

sec\ \varnothing = \frac{r}{x}

sec\ \varnothing = \frac{5\sqrt 2}{-7}

sec\ \varnothing = \frac{-5}{7}\sqrt 2

csc\ \varnothing = \frac{r}{y}

csc\ \varnothing = \frac{5\sqrt 2}{-1}

csc\ \varnothing = -5\sqrt 2

3 0
3 years ago
Round to the near at tenth <br> | 3x + 1 | =22
sdas [7]

Answer:

x =7                                    x = -7  2/3

Step-by-step explanation:

| 3x + 1 | =22

Absolute value equations have 2 solutions, one positive and one negative

3x+1 = 22                   3x+1 = -22

Subtract 1 from each side

3x+1-1 =22-1               3x+1-1 = -22-1

3x= 21                         3x = -23

Divide each side by 3

3x/3 = 21/3                      3x/3 = -23/3

x =7                                    x = -7  2/3

6 0
3 years ago
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