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Ber [7]
3 years ago
15

5. Consider the diagram below. Solve for x.

Mathematics
1 answer:
mrs_skeptik [129]3 years ago
4 0

Answer:

41 degrees

Step-by-step explanation:

The three angles need to add up to 180 degrees!

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Which equation is equivalent to log Subscript 3 Baseline (x + 5) = 2?
adell [148]

Answer:

The correct option is right-bracket squared 3 squared =x+5

Step-by-step explanation:

The equation is \log _{3}(x+5)=2

Option a: \log _{3}(x+5)=3^{2}

This is not possible because using logarithmic rule, if \log _{a} b=c then b=a^{c}

Hence, option a is not equivalent to \log _{3}(x+5)=2

Option b: \log _{3}(x+5)=2^{3}

This is not possible because using logarithmic rule, if \log _{a} b=c then b=a^{c}

Hence, option b is not equivalent to \log _{3}(x+5)=2

Option c: x+5=3^{2}

This is possible because using logarithmic rule, if \log _{a} b=c then b=a^{c}

Hence, option c is equivalent to \log _{3}(x+5)=2

Option b: x+5=2^{3}

This is not possible because using logarithmic rule, if \log _{a} b=c then b=a^{c}

Hence, option b is not equivalent to \log _{3}(x+5)=2

Thus, the correct option is c: x+5=3^{2}

Hence, the equation x+5=3^{2} is equivalent to \log _{3}(x+5)=2

8 0
3 years ago
C) Write 2400 000 in standard form.
svp [43]

Answer:

2 400 000

Step-by-step explanation:

divide the numbers into 3

8 0
3 years ago
Read 2 more answers
SOLVE WITH AN EXPLANATION PLS
Softa [21]

Answer:

  • Option C
  • Option B

Step-by-step explanation:

<u>Let's simplify this further to get our answer.</u>

  • -5/6(12 - 6x + 18y)
  • => -60/6 + 30x/6 - 90y/6
  • => -10 + 5x - 15y

Looking at the options, we can say that Option C and B are correct.

5 0
2 years ago
Read 2 more answers
Ms. Bedru conducts a survey of the ages
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5 0
2 years ago
Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3
omeli [17]
\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}

=\frac{(tanx+cotx)^2+(tan^2x-cot^2x)^2}{(tanx+cotx)^2}=\frac{(tanx+cotx)^2}{(tanx+cotx)^2}+\frac{(tan^2x-cot^2x)^2}{(tanx+cotx)^2}\\\\=1+\frac{(tanx-cotx)^2(tanx+cotx)^2}{(tanx+cotx)^2}=1+(tanx-cotx)^2\\\\=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\\\\=\left(\frac{sinx}{cosx}\right)^2+\left(\frac{cosx}{sinx}\right)^2-1=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-1=\frac{sin^4x+cos^4x}{sin^2x\ cos^2x}-1

=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R
3 0
3 years ago
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