1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Gemiola [76]
3 years ago
15

Approximately what portion of the box is shaded blue? A.3/5 B.2/3 C.9/10

Mathematics
1 answer:
BaLLatris [955]3 years ago
6 0

Answer:

I think the answer is C.....

You might be interested in
Steven bought treats for his friends at the mall food court. Where m,n,p, and q are whole numbers, the amount Steven spent on th
Fed [463]

We have that the term from the expression that represents the amount of money Steven spent on cinnamon rolls is

X=3.19p

From the question we are told that

m,n,p, and q are whole numbers,

Expression Total amount spent =2.35m + 2.75n + 3.19p + 3.39q

Cost of cinnamon rolls=\$3.19

Generally the equation for the Money spent on cinnamon roll   is mathematically given as

X=No of cinnamon roll * Cost of cinnamon roll

X=3.19*p

X=3.19p

For more information on this visit

brainly.com/question/23366835?referrer=searchResults

7 0
3 years ago
(1,-4), (2,1), (3,4), (3,3), (4,2) is this a function?
djverab [1.8K]

no i think because theres a negative

3 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} } }

<u>Solving the given expression</u><u> </u><u>:</u><u>-</u><u> </u>

\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)}

<u>Using</u><u> </u><u>(</u><u>a</u><u> </u><u>+</u><u> </u><u>b</u><u> </u><u>)</u><u> </u><u>(</u><u>a</u><u> </u><u>-</u><u> </u><u>b</u><u> </u><u>)</u><u> </u><u>=</u><u> </u><u>a²</u><u> </u><u>-</u><u> </u><u>b²</u>

\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
On a piece of paper, use a protractor to construct a triangle with angle measures of 40° and 60º.
andrew-mc [135]

Answer:

There are 180 degrees in a triangle so it is 80 degrees

4 0
2 years ago
Read 2 more answers
What is the mathematical relationship between the atomic radius (r) and the lattice parameter (a0) in the BCC structure? a. b. c
ryzh [129]

Answer: Lattice parameter, a = (4R)/(√3)

Step-by-step explanation:

The typical arrangement of atoms in a unit cell of BCC is shown in the first attachment.

The second attachment shows how to obtain the value of the diagonal of the base of the unit cell.

If the diagonal of the base of the unit cell = x

(a^2) + (a^2) = (x^2)

x = a(√2)

Then, diagonal across the unit cell (a cube) makes a right angled triangle with one side of the unit cell & the diagonal on the base of the unit cell.

Let the diagonal across the cube be y

Pythagoras theorem,

(a^2) + ((a(√2))^2) = (y^2)

(a^2) + 2(a^2) = (y^2) = 3(a^2)

y = a√3

But the diagonal through the cube = 4R (evident from the image in the first attachment)

y = 4R = a√3

a = (4R)/(√3)

QED!!!

8 0
3 years ago
Read 2 more answers
Other questions:
  • what is the Best way to do math problem like geometry, angle problems, circumference and other things like that ?
    9·2 answers
  • As indicated below, write the equations of the line passing through the point (2, 4) and parallel to the line whose equation is
    15·1 answer
  • What is 2 2/5 as a decimal?
    12·2 answers
  • I need help with this question
    13·2 answers
  • PLEASE HELP BRAINLIEST!!!!
    9·2 answers
  • Enter the unknown number that makes the equation true. 75 + 3 x ☐ = 372
    11·2 answers
  • There are a total of 66 people attending a carnival. The ratio of adults to children is 4 to 7. How many adults are attending th
    9·1 answer
  • What is the value of p in this equation 10p – 8 ≥ 12
    8·2 answers
  • (05.06 LC)
    12·1 answer
  • Simplify 1/6^3 divided by 1/36
    12·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!