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3 years ago

the sum of 3 decimal numbers is 6. Exactly one of the numbers is less than 1 what could those numbers be.

1 answer:

3 0

There are infinitely many solutions to this problem (which is why it asks what could those numbers be), so you just have to make one up!

Start with the first sentence.
x + y + z = 6.0

Now one of those numbers has to be less than 1. I'll choose x. And I'll choose a random number, say, 0.7.

0.7 + y + z = 6.0

Now I'll subtract 0.7 from both sides to see what y and z have to add up to.

0.7 + y + z - 0.7 = 6.0 - 0.7
y + z = 5.3

Now I'll choose a random number for y that's less than 5.3.

2.6 + z = 5.3

And subtract 2.6 from both sides to find z.

2.6 + z - 2.6 = 5.3 - 2.6
z = 2.7

And now we have our numbers!
x = 0.7
y = 2.6
z = 2.7

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A box with a square base and open top must have a volume of 296352 c m 3 . We wish to find the dimensions of the box that minimi

mestny [16]

Answer:

Base Length of 84cm
Height of 42 cm.

Step-by-step explanation:

Given a box with a square base and an open top which must have a volume of 296352 cubic centimetre. We want to minimize the amount of material used.

Step 1:

Let the side length of the base =x

Let the height of the box =h

Since the box has a square base

Volume, $V=x^2h=296352$

$h=\dfrac{296352}{x^2}$

Surface Area of the box = Base Area + Area of 4 sides

$A(x,h)=x^2+4xh\\$Substitute h=\dfrac{296352}{x^2}\\A(x)=x^2+4x\left(\dfrac{296352}{x^2}\right)\\A(x)=\dfrac{x^3+1185408}{x}$

Step 2: Find the derivative of A(x)

$If\:A(x)=\dfrac{x^3+1185408}{x}\\A'(x)=\dfrac{2x^3-1185408}{x^2}$

Step 3: Set A'(x)=0 and solve for x

$A'(x)=\dfrac{2x^3-1185408}{x^2}=0\\2x^3-1185408=0\\2x^3=1185408\\$Divide both sides by 2\\x^3=592704\\$Take the cube root of both sides\\x=\sqrt[3]{592704}\\x=84$

Step 4: Verify that x=84 is a minimum value

We use the second derivative test

$A''(x)=\dfrac{2x^3+2370816}{x^3}\\$When x=84$\\A''(x)=6$

Since the second derivative is positive at x=84, then it is a minimum point.

Recall:

$h=\dfrac{296352}{x^2}=\dfrac{296352}{84^2}=42$

Therefore, the dimensions that minimizes the box surface area are:

Base Length of 84cm
Height of 42 cm.

5 0

2 years ago

Emie ordered 3 buckets of fried chicken at $11.95 each 5 pints and Cole slaw at $2.19 each. The sales tax was $3.04. About how m

Oliga [24]

Answer:

Around $49 the actual answer is $49.84

3 0

2 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 :-

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