All categories

3 years ago

Expand and simplity 4(2x - 1)3(2x 5) Plzz help me

Mathematics

1 answer:

k0ka [10]3 years ago

7 0

Answer:

240^2 - 120x

Step-by-step explanation:

4(2x - 1) x 3 x (2x 5)

4(2x - 1) x 3 x (10x)

4(2x - 1) x 3 x 10x

120x x (2x-1)

240^2 - 120x

You might be interested in

BRAINLIEST AND POINTS!! EASY!!!

SVETLANKA909090 [29]

Hey there!

The mean of the numbers is the average. To find the mean we will simply add all the numbers together, and divide them by the amount of numbers there are. Easy, peesy, lemon squeezy. Lets get started.

7 + 10 + 22 + 10 + 12 + 10 + 12 = 83

Now that we have added those altogether, we can divide them by the amount of numbers there are. The amount of numbers there are comes to a total of 7. Therefore...

83 divided by 7 = 11.857

Answer Choice B. 11.9 is your correct answer.

3 0

3 years ago

Read 2 more answers

Let measure angle A equal 40. If angle B complement of angle A, and angle C is a supplemlement of angle B find these measures

uranmaximum [27]

Two angles are complementary when they add up to 90 degrees
m∠B = 90° - m∠A = 90° - 40° = 50°

Two angles are supplementary when they add up to 180 degrees
m∠C = 180° - m∠B = 180° - 50° = 130°

8 0

3 years ago

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

2 years ago

A swimming pool has a capacity of 2,500 gallons of water. The output of a hose is 25 gallon per minute. How much time would it t

g100num [7]

Answer:

100 minutes

Solution:

We know that:

Capacity = 2,500 gallons
Hose = 25 gallons/minute

Solution:

25 gallons = 1 minute
=> 25x gallons = x minutes = 2500 gallons
=> 25x = 2500
=> x = 100 minutes

Hence, it will take 100 minutes to fill the pool.

4 0

3 years ago

Find the empirical formula of the compound containing 2.4 gram of carbon, 6.4 gram of oxygen and 0.2 gram of hydrogen

Usimov [2.4K]

Answer:

Empirical formula= COOH

Step-by-step explanation:

Molecular mass of the elements

Carbon= 12

Oxygen= 16

Hydrogen= 1

We divide the elements each with their molecular formula

Carbon= 2.4/12

Carbon= 0.2

Oxygen= 6.4/16

Oxygen= 0.4

Hydrogen= 0.2/1

Hydrogen= 0.2

Now we divide with the smallest result which is 0.2

Carbon= 0.2/0.2

Carbon = 1

Oxygen= 0.4/0.2

Oxygen= 2

Hydrogen= 0.2/0.2

Hydrogen= 1

So we have

Carbon 1, oxygen 2, hydrogen 1

Empirical formula= COOH

4 0

3 years ago