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2 years ago

3. State if each statement is correct. If it is not correct, then fix the mistake.

Mathematics

1 answer:

olya-2409 [2.1K]2 years ago

4 0

Answer:

All are incorrect

Step-by-step explanation:

Required

Determine which is correct or not

$(a)\ 2x * 2x = 4x$

This is incorrect because:

$2x * 2x = 2 * 2 * x*x$

$2x * 2x = 4x^2$ ---- fixed

$(b)\ 3x^3 * 3x^3 = 9x^9$

This is incorrect because:

$3x^3 * 3x^3 = 3 * 3 * x^3 * x^3$

$3x^3 * 3x^3 = 9x^(3+3)$

$3x^3 * 3x^3 = 9x^6$ ---- fixed

$(c)\ 4x(4x + 5) =8x + 5$

This is incorrect because:

$4x(4x + 5) =4x * 4x + 4x * 5$

$4x(4x + 5) =16x^2 + 20x$ ---- fixed

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Answer:

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Step-by-step explanation:

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1 year ago

Help me please ASAP. I really need this

Ksju [112]

Answer:

Step-by-step explanation:

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2 years ago

I don’t understand the question

kipiarov [429]

Answer:

Answer is given below with explanations.

Step-by-step explanation:

$- 13 {x}^{2} \: has \: only \: one \: term \: and \: highest \: degree \: is \: 2 \\$

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<em>HAVE A NICE DAY</em><em>!</em>

<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em>.</em>

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2 years ago

Round to nearest hundred 64,505

NARA [144]

64,500 would be the answer

7 0

2 years ago

Read 2 more answers

Find the measures of the three angles, in radians, of the triangle with the given vertices: d(1,1,1), e(1,−5,2), and f(−2,2,7).

Oduvanchick [21]

Consider triangle DEF with vertices D(1,1,1), E(1,-5,2) and F(-2,2,7).

1. Find

$\overrightarrow{DE}=(1-1,-5-1,2-1)=(0,-6,1),\\ \\\overrightarrow{DF}=(-2-1,2-1,7-1)=(-3,1,6).$

Then

$\cos \angle D=\dfrac{0\cdot (-3)+(-6)\cdot 1+1\cdot 6}{\sqrt{0^2+(-6)^2+1^2}\cdot \sqrt{(-3)^2+1^2+6^2}}=\dfrac{0}{\sqrt{37} \cdot \sqrt{46} }=0.$

2. Find

$\overrightarrow{ED}=(1-1,1-(-5),1-2)=(0,6,-1),\\ \\\overrightarrow{EF}=(-2-1,2-(-5),7-2)=(-3,7,5).$

Then

$\cos \angle E=\dfrac{0\cdot (-2)+6\cdot 7+(-1)\cdot 5}{\sqrt{0^2+6^2+(-1)^2}\cdot \sqrt{(-3)^2+7^2+5^2}}=\dfrac{37}{\sqrt{37} \cdot \sqrt{83} }=\sqrt{\dfrac{37}{83}}.$

3. Find

$\overrightarrow{FE}=(1-(-2),-5-2,2-7)=(3,-7,-5),\\ \\\overrightarrow{FD}=(1-(-2),1-2,1-7)=(3,-1,-6).$

Then

$\cos \angle F=\dfrac{3\cdot 3+(-7)\cdot (-1)+(-5)\cdot (-6)}{\sqrt{3^2+(-7)^2+(-5)^2}\cdot \sqrt{3^2+(-1)^2+(-6)^2}}=\dfrac{46}{\sqrt{83} \cdot \sqrt{46} }=\sqrt{\dfrac{46}{83}}.$

$\angle E=\arccos0=\dfrac{\pi}{2},\\ \\
\angle D=\arccos\letf(\sqrt{\dfrac{37}{83}}\right)\approx 0.27\pi,\\ \\
\angle F=\arccos\letf(\sqrt{\dfrac{46}{83}}\right)\approx 0.23\pi.$

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2 years ago