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4 years ago

How do you find the missing side lengths?

Mathematics

1 answer:

svet-max [94.6K]4 years ago

3 0

Since this is a right triangle, one of the angles is going to be 90. And just with the other two angles would have to add up to 90

Example: One angle is 90 Degrees. Another angle is 30 Degrees and the last angle has to be 60, because 30+60=90

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Where is the mistake in the following work? *

stealth61 [152]

Step-by-step explanation:

step 2
$- 9 = - 3x + 5 + 13 \\ - 9 = 2x + 13 \\ - 2x = 4 \\ - 2x \div - 2 = 4 \div - 2 \\ x = - 2$

8 0

3 years ago

PLZ HELP QUICK<br> Which value from the set {12, 16, 21, 36} makes 2/3-n=19 true?

lilavasa [31]

Answer: 36

Step-by-step explanation: when you solve the problem 2/3n-5=19 you get 36.

3 0

3 years ago

Read 2 more answers

PLEASE HELP ITS MULTIPLE CHOICE i added extra points and i’ll mark BRAINLIEST !!!

Artemon [7]

Answer:

Step-by-step explanation:

3 0

3 years ago

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Use mathematical induction to prove that for each integer n > 4,5" > 2^2n+1 + 100.

Flura [38]

Answer:

The inequality that you have is $5^{n}>2^{2n+1}+100,\,n>4$ . You can use mathematical induction as follows:

Step-by-step explanation:

For $n=5$ we have:

$5^{5}=3125$

$2^{(2(5)+1)}+100=2148$

Hence, we have that $5^{5}>2^{(2(5)+1)}+100.$

Now suppose that the inequality holds for $n=k$ and let's proof that the same holds for $n=k+1$ . In fact,

$5^{k+1}=5^{k}\cdot 5>(2^{2k+1}+100)\cdot 5.$

Where the last inequality holds by the induction hypothesis.Then,

$5^{k+1}>(2^{2k+1}+100)\cdot (4+1)$

$5^{k+1}>2^{2k+1}\cdot 4+100\cdot 4+2^{2k+1}+100$

$5^{k+1}>2^{2k+3}+100\cdot 4$

$5^{k+1}>2^{2(k+1)+1}+100$

Then, the inequality is True whenever $n>4$ .

3 0

3 years ago

Simplify (12a5−6a−10a3)−(10a−2a5−14a4) . Write the answer in standard form

almond37 [142]

Answer:

$=14a^5+14a^4-10a^3-16a$

Step-by-step explanation:

$\left(12a^5-6a-10a^3\right)-\left(10a-2a^5-14a^4\right)\\\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\\=12a^5-6a-10a^3-\left(10a-2a^5-14a^4\right)\\-\left(10a-2a^5-14a^4\right):\quad -10a+2a^5+14a^4\\-\left(10a-2a^5-14a^4\right)\\\mathrm{Distribute\:parentheses}\\=-\left(10a\right)-\left(-2a^5\right)-\left(-14a^4\right)\\Apply\:minus-plus\:rules\\-\left(-a\right)=a,\:\:\:-\left(a\right)=-a\\=-10a+2a^5+14a^4\\=12a^5-6a-10a^3-10a+2a^5+14a^4$

$\mathrm{Simplify}\:12a^5-6a-10a^3-10a+2a^5+14a^4:\quad 14a^5+14a^4-10a^3-16a12a^5-6a-10a^3-10a+2a^5+14a^4\\Group\:like\:terms\\=12a^5+2a^5+14a^4-10a^3-6a-10a\\\mathrm{Add\:similar\:elements:}\:12a^5+2a^5=14a^5\\=14a^5+14a^4-10a^3-6a-10a\\\mathrm{Add\:similar\:elements:}\:-6a-10a=-16a\\=14a^5+14a^4-10a^3-16a$

4 0

3 years ago