Ask question

Ask question

All categories

antoniya [11.8K]

3 years ago

5

. The cost of petrol rises by 2 cents a liter. last week a man bought 20 liters at the old price. This week he bought 10 liters

at the new price. Altogether, the petrol costs $9.20. What was the old price for 1 liter?

2 answers:

xeze [42]3 years ago

4 0

0.46 maybe you should ask your teacher for help more then she can help and maybe you would know more then everyone

NNADVOKAT [17]3 years ago

3 0

Answer:

0.46

Step-by-step explanation:

You might be interested in

Which postulate or theorem would you use to prove the triangles congruent?; Which pair of triangles Cannot be proved congruent?;

Lunna [17]

Proving triangles are congruent triangles uses theorems (postulates), the Angle Side Angle (ASA), Side Angle Side (SAS), Side Side Side (SSS), and AAS (Angle-Angle-Side). Two triangles with equal corresponding angles may not be congruent to each other because one triangle might be an enlarged copy of the other.

Triangles that have exactly the same size and shape are called congruent triangles. The congruent triangles can be used using the Angle Side Angle (ASA), Side Angle Side (SAS), Side Side Side (SSS), and AAS (Angle-Angle-Side) theorems.

Angle Side Angle (ASA) theorem states that if any two angles and the side included between the angles of one triangle are equivalent to the corresponding two angles and side included between the angles of the second triangle, then the two triangles are said to be congruent.

Side-angle-side (SAS) theorem states that if two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, the triangles are congruent.

Side Side Side (SSS) theorem states that if all the three sides of one triangle are equivalent to the corresponding three sides of the second triangle, then the two triangles are said to be congruent.

AAS (Angle-Angle-Side) theorem states that if two angles and a non-included side of a triangle are equal to the corresponding angles and sides of another triangle, then the triangles are said to be congruent.

Learn more about Congruent triangles:

brainly.com/question/1675117

#SPJ4

6 0

1 year ago

If the outliers are not included what is the mean of the data set, 79 79 80 82 50 78 83 79 81 82

White raven [17]

The mean to your question is 62.3
you need to add all of the numbers together and then divide by how many numbers are there to find the answer.

3 0

2 years ago

Given a dilation around the origin, what is the scale factor K?

kobusy [5.1K]

Answer:

The answer is " $\frac{4}{3}$ "

Step-by-step explanation:

Given:

$\bold{DoK = (12, 8) (9,6)}$

In the scale factor when we take x coordinate then the value of k is:

$\Rightarrow x= \frac{x_1}{x_2}\\\\\Rightarrow x= \frac{12}{9}\\\\\Rightarrow x= \frac{4}{3}\\$

If we take y coordinate then the value of k is:

$\Rightarrow y= \frac{y_1}{y_2}\\\\\Rightarrow y= \frac{8}{6}\\\\\Rightarrow y= \frac{4}{3}\\$

So, the final value of k is " $\frac{4}{3}$ " .

5 0

2 years ago

Which of the following equations have no solutions? Choose all answers that apply:

Marianna [84]

Answer:

All

Step-by-step explanation:

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 $\sf \theta$

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

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

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

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

<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