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2 years ago

How do you solve (2y^2)^3 x 2y^3

Mathematics

2 answers:

andrew-mc [135]2 years ago

8 0

Answer:

8y^6,2xy^3

Step-by-step explanation:

balu736 [363]2 years ago

8 0

Hope it helps !!!!!!!!!!!

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Please answer this question, i request

Jet001 [13]

${\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}$

$\star \: \tt \cot \theta = \dfrac{7}{8}$

${\large{\textsf{\textbf{\underline{\underline{To \: Evaluate :}}}}}}$

$\star \: \tt \dfrac{(1 + \sin \theta)(1 - \sin \theta) }{(1 + \cos \theta) (1 - \cos \theta) }$

${\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}$

Consider a $\triangle$ ABC right angled at C and $\sf \angle \: B = \theta$

Then,

‣ Base [B] = BC

‣ Perpendicular [P] = AC

‣ Hypotenuse [H] = AB

$\therefore \tt \cot \theta = \dfrac{Base}{ Perpendicular} = \dfrac{BC}{AC} = \dfrac{7}{8}$

Let,

Base = 7k and Perpendicular = 8k, where k is any positive integer

In $\triangle$ ABC, H² = B² + P² by Pythagoras theorem

$\longrightarrow \tt {AB}^{2} = {BC}^{2} + {AC}^{2}$

$\longrightarrow \tt {AB}^{2} = {(7k)}^{2} + {(8k)}^{2}$

$\longrightarrow \tt {AB}^{2} = 49{k}^{2} + 64{k}^{2}$

$\longrightarrow \tt {AB}^{2} = 113{k}^{2}$

$\longrightarrow \tt AB = \sqrt{113 {k}^{2} }$

$\longrightarrow \tt AB = \red{ \sqrt{113} \: k}$

Calculating Sin $\sf \theta$

$\longrightarrow \tt \sin \theta = \dfrac{Perpendicular}{Hypotenuse}$

$\longrightarrow \tt \sin \theta = \dfrac{AC}{AB}$

$\longrightarrow \tt \sin \theta = \dfrac{8 \cancel{k}}{ \sqrt{113} \: \cancel{ k } }$

$\longrightarrow \tt \sin \theta = \purple{ \dfrac{8}{ \sqrt{113} } }$

Calculating Cos $\sf \theta$

$\longrightarrow \tt \cos \theta = \dfrac{Base}{Hypotenuse}$

$\longrightarrow \tt \cos \theta = \dfrac{BC}{ AB}$

$\longrightarrow \tt \cos \theta = \dfrac{7 \cancel{k}}{ \sqrt{113} \: \cancel{k } }$

$\longrightarrow \tt \cos \theta = \purple{ \dfrac{7}{ \sqrt{113} } }$

Solving the given expression :- 

$\longrightarrow \: \tt \dfrac{(1 + \sin \theta)(1 - \sin \theta) }{(1 + \cos \theta) (1 - \cos \theta) }$

Putting,

• Sin $\sf \theta$ = $\dfrac{8}{ \sqrt{113} }$

• Cos $\sf \theta$ = $\dfrac{7}{ \sqrt{113} }$

$\longrightarrow \: \tt \dfrac{ \bigg(1 + \dfrac{8}{ \sqrt{133}} \bigg) \bigg(1 - \dfrac{8}{ \sqrt{133}} \bigg) }{\bigg(1 + \dfrac{7}{ \sqrt{133}} \bigg) \bigg(1 - \dfrac{7}{ \sqrt{133}} \bigg)}$

Using (a + b ) (a - b ) = a² - b²

$\longrightarrow \: \tt \dfrac{ { \bigg(1 \bigg)}^{2} - { \bigg( \dfrac{8}{ \sqrt{133} } \bigg)}^{2} }{ { \bigg(1 \bigg)}^{2} - { \bigg( \dfrac{7}{ \sqrt{133} } \bigg)}^{2} }$

$\longrightarrow \: \tt \dfrac{1 - \dfrac{64}{113} }{ 1 - \dfrac{49}{113} }$

$\longrightarrow \: \tt \dfrac{ \dfrac{113 - 64}{113} }{ \dfrac{113 - 49}{113} }$

$\longrightarrow \: \tt { \dfrac { \dfrac{49}{113} }{ \dfrac{64}{113} } }$

$\longrightarrow \: \tt { \dfrac{49}{113} }÷{ \dfrac{64}{113} }$

$\longrightarrow \: \tt \dfrac{49}{ \cancel{113}} \times \dfrac{ \cancel{113}}{64}$

$\longrightarrow \: \tt \dfrac{49}{64}$

$\qquad \: \therefore \: \tt \dfrac{(1 + \sin \theta)(1 - \sin \theta) }{(1 + \cos \theta) (1 - \cos \theta) } = \pink{\dfrac{49}{64} }$

$\begin{gathered} {\underline{\rule{300pt}{4pt}}} \end{gathered}$

${\large{\textsf{\textbf{\underline{\underline{We \: know :}}}}}}$

✧ Basic Formulas of Trigonometry is given by :-

$\begin{gathered}\begin{gathered}\boxed { \begin{array}{c c} \\ \bigstar \: \sf{ In \:a \:Right \:Angled \: Triangle :} \\ \\ \sf {\star Sin \theta = \dfrac{Perpendicular}{Hypotenuse}} \\\\ \sf{ \star \cos \theta = \dfrac{ Base }{Hypotenuse}}\\\\ \sf{\star \tan \theta = \dfrac{Perpendicular}{Base}}\\\\ \sf{\star \cosec \theta = \dfrac{Hypotenuse}{Perpendicular}} \\\\ \sf{\star \sec \theta = \dfrac{Hypotenuse}{Base}}\\\\ \sf{\star \cot \theta = \dfrac{Base}{Perpendicular}} \end{array}}\\\end{gathered} \end{gathered}$

${\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}$

✧ Figure in attachment

$\begin{gathered} {\underline{\rule{200pt}{1pt}}} \end{gathered}$

3 0

1 year ago

What is the value of x to the nearest tenth? 3.1 4.8 5.4 3.5

Tanya [424]

Answer:

5.4

Step-by-step explanation:

I am not really sure about this type of problem, but a diameter has to cross the midpoint of the circle the dark dot and hit two point on both side of the circle. Yet the line 9.6 is not a diameter since it didn't go through the center, so the diameter has to be longer than 9.6. But x is looking for radius. so divide 9.6 by 2 u get 4.8 but that is not a diameter and the real diameter is longer that mean the radius is longer so the only answer is 5.4

8 0

3 years ago

Help it’s confusing

noname [10]

Answer:

I think is C like I said not quite sure...

8 0

2 years ago

Read 2 more answers

If Rick wishes to reduce his BMI to 27, he needs to eat fewer kcalories than he expends. For an adolescent who carries excess fa

Nesterboy [21]

This question is not complete because his height and age was not indicated in the above question.

Complete Question:

Rick is a healthy 19-year-old college student who is 70 inches tall and weighs 205 pounds. He has decided to "get a six-pack" over the summer with a diet and exercise program. As part of his new plan, he has stopped drinking soda and is eating more salads in addition to his usual diet. Besides of these changes, he is unclear on how to proceed to reach his fitness goal. Rick's mother wants to make sure his approach will not interfere with his normal growth and development, and has asked him to seek reliable information to help him make a reasonable plan.

If Rick wishes to reduce his BMI to 27, he needs to eat fewer kcalories than he expends. For an adolescent who carries excess fat, the recommended maximal weight loss is one pound per week. Since there are 3500 kcalories in a pound of body fat, a deficit of 3500 kcalories for the week or 500 kcalories per day would be required. Calculate the maximum number of kcalories Rick can consume per day to achieve a weight loss of one pound per week. Assume that his weight is 205 pounds and that his physical activity factor is "low active."

Answer:

2696 kilocalories

Step-by-step explanation:

STEP 1

First we need to calculate Rick's Basal Metabolic Rate( BMR)

Weight in pounds = 205 pounds

Age = 19 years

Height in inches = 70 inches

The formula for calculating Basal Metabolic Rate for men =

BMR = 66.47 + ( 6.24 × weight in pounds ) + ( 12.7 × height in inches ) − ( 6.755 × age in years )

BMR = 66.47 + ( 6.24 × 205 ) + ( 12.7 × 70 ) − ( 6.755 × 19)

= 2247

STEP 2:

The next step is to calculate the maximum amount kilocalories per day Rick should consume

Formula is given as :

For a person with physical activity factor of a ' low Active'

Maximum amount of kilocalories to consume per day = Basal Metabolic rate × 1.2

= 2247 × 1.2

= 2696 Kilocalories (kcalories) per day

6 0

2 years ago

RST is an equilateral triangle; If TX = 6 then RX =

lara31 [8.8K]

In the figure, tx is perpendicular to line RS.this means we can use the pythagorean theorem to determine RX length the long way or simply apply the theorem that each divided triangle becomes a 30-60-90 angle. If opposite to 60 deg is equal to 6 units, then the side's length is equal to 2*6/sq rt 3 or equal to 6.93 units. The answer is half of this equal to 3.47 units.

6 0

3 years ago

Read 2 more answers