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3 years ago

I need some help please

Mathematics

2 answers:

Juliette [100K]3 years ago

7 0

My Answer for this is I think 77 for x

Naily [24]3 years ago

6 0

Answer:

Step-by-step explanation:

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I know the selected answer is correct but I'm not too sure how to get that answer.

Kryger [21]

$\tt{ Hey \: there , \: Mr.Panda \: ! }$ ;)

♨ $\large{ \tt{ E \: X \: P \: L \: A \: N \: A \: T \: I\: O \: N}}:$

⤻ Before solving the given question , you should know the answer of these questions :

✺How do you find the hypotenuse , perpendicular and base when the angle ( $\theta \: , \alpha \: ,\beta$ ) is given ?

⇾ The longest side , which is the opposite side of right angle is the hypotenuse ( h ). There are two other sides , the opposite and the adjacent. The naming of these sides depends upon which angle is involved. The opposite is the side opposite the angle involved and it is called the perpendicular ( p ) . The adjacent us the side next to the angle involved ( buy not the hypotenuse ) and it is called the base ( b ).

☄ $\large{ \tt{REMEMBER}} :$

$\bf{ \sin \theta = \frac{opposite}{hypotenuse} = \frac{perpendicular}{hypotenuse} }$

$\bf{ \cos\theta = \frac{adjacent}{hypotenuse} = \frac{base}{hypotenuse} }$

$\bf{ \tan \theta = \frac{opposite}{adjacent} = \frac{perpendicular}{base} }$

In the above cases , $\theta$ is taken as the angle of reference.

♪ Our Q/A part ends up here! Let's start solving the question :

❈ $\large{ \tt{GIVEN}} :$

Perpendicular ( p ) = ? , Hypotenuse ( h ) = 18 & base ( b ) = 16

✧ $\large{ \tt{TO \: FIND} : }$

Value of tan $\theta$

✎ $\large{ \tt{SOLUTION}} :$

Firstly , Finding the value of perpendicular ( p ) using Pythagoras theorem :

❃ $\boxed{ \sf{ {h}^{2} = {p}^{2} + {b}^{2} }}$ [ Pythagoras theorem ]

$\large{ ⇢ \sf{p}^{2} + {b}^{2} = {h}^{2} }$

$\large{⇢ \sf{ {p}^{2} = {h}^{2} - {b}^{2} }}$

$\large{ ⇢\sf{ {p}^{2} = {18}^{2} - {16}^{2} }}$

$\large{⇢ \sf{ {p}^{2} = 324 - 256}}$

$\large{⇢ \sf{ {p}^{2} = 68}}$

$\large{⇢ \sf{p = \sqrt{68}}}$

$\large{ ⇢\sf{p = \boxed{ \tt{2 \sqrt{17}}} }}$

Okey, We found out the perpendicular i.e $\tt{2 \sqrt{17}}$ . Now , We know :

❊ $\large{ \sf{ \tan \theta} = \frac{perpendicular}{base} }$

$\large {\tt{↬ \: tan \theta = \frac{2 \sqrt{17} }{16}}}$

$\large{ \tt{ ↬ tan \theta = \frac{ \cancel{2} \: \sqrt{17} }{ \cancel{16} \: \: 8} }}$

$\large{ \tt{ ↬ \boxed{ \tt{tan \theta = \frac{ \sqrt{17} }{8}}}}}$

⟿ $\boxed{ \boxed{ \tt{OUR\: FINAL \: ANSWER : \boxed{ \underline{ \bf{ \frac{ \sqrt{17} }{8}}}}}}}$

۵ Yay! We're done!

♕ $\large\tt{RULE \: OF \:SUCCESS }:$

Never lose hope & keep on working ! ✔

ツ Hope I helped!

☃ Have a wonderful day / evening! ☼

# StayInAndExplore ☂

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3 0

3 years ago

Read 2 more answers

PLZ HELP 67 POINTS. 1 + 1<br>

Arada [10]

Answer:

100% its 2 my guy trust meeee

7 0

3 years ago

Read 2 more answers

Use the data below to construct a steam and leaf display on your own paper, then describe the distribution's shape

Art [367]

Good job <span>Use the data below to construct a steam and leaf display on your own paper, then describe the distribution's shape</span>

4 0

3 years ago

A number plus the sum of the number<br> and seven.

Dmitrij [34]

Answer:

x+(x+7)

Step-by-step explanation:

sum=add all together

8 0

3 years ago

A jeweler wants to make 14 grams of an alloy that is precisely 75% gold.. The jeweler has alloys that are 25% gold, 50% gold, &a

Goryan [66]

Given that the jeweler has alloys that are 25% gold, 50% gold, and 82% gold.

As he wants to make 14 grams of an alloy by adding two different alloys that is precisely 75% gold, so one alloy must have a percentage of gold more than 75%.

One alloy is 82% gold and, the second can be chosen between 25% gold, 50% gold, so there are two cases.

Case 1: 82% gold + 50% gold

Let x grams of 82% gold and y grams of 50% gold added to make x+y=14 grams of 75% gold, so

75% of 14 = 82% of x + 50% of y

$\Rightarrow 75/100 \times 14 = 82/100 \times x + 50/100 \times y \\\\$

$\Rightarrow 75/100 \times 14 = 82/100 \times x + 50/100 \times (14-x)$ [as x+y=14]

$\Rightarrow 75 \times 14 = 82 \times x + 50 \times (14-x) \\\\\Rightarrow 75 \times 14 = 82 \times x + 50 \times14-50\times x \\\\\Rightarrow 75 \times 14 = 32 \times x + 50 \times14 \\\\\Rightarrow 32 \times x =75 \times 14 - 50 \times14 \\\\$

$\Rightarrow x =(25 \times 14)/32=10.9375$ grams

and $y = 14-x= 14-10.9375=3.0625$ grams.

Hence, 10.9375 grams of 82% gold and 3.0625 grams of 50% gold added to make 14 grams of 75% gold.

Case 2: 82% gold + 25% gold

Let x grams of 82% gold and y grams of 25% gold added to make x+y=14 grams of 75% gold, so

75% of 14 = 82% of x + 25% of y

$\Rightarrow 75/100 \times 14 = 82/100 \times x + 25/100 \times y \\\\\Rightarrow 75/100 \times 14 = 82/100 \times x + 25/100 \times (14-x) \\\\ \Rightarrow 75 \times 14 = 82 \times x + 25 \times (14-x) \\\\\Rightarrow 75 \times 14 = 82 \times x + 25 \times14-25\times x \\\\\Rightarrow 75 \times 14 = 57 \times x + 25 \times14 \\\\\Rightarrow 57 \times x =75 \times 14 - 25 \times14 \\\\$

$\Rightarrow x =(50 \times 14)/57=12.28$ grams

and $y = 14-x= 14-12.28=1.72$ grams.

Hence, 12.28 grams of 82% gold and 1.72 grams of 50% gold added to make 14 grams of 75% gold.

3 0

2 years ago