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hram777 [196]
3 years ago
10

What is the value of y? Enter your answer, as an exact value, in the box.

Mathematics
1 answer:
OLga [1]3 years ago
7 0
We know the bottom triangle is a 45, 45, 90 triangle, so the hypotenuse is √2 times the value of the legs:

(√2)(√2)
=√4
=2

Now, we can use this to solve for y. The top triangle is a 30, 60, 90 triangle. The side we found above is the side across from the 30 degree angle. The side opposite the 60 degree angle is √3 times the side across the 30 degree angle. Therefore, we can solve for y by multiplying 2 by √3

y=2√3
You might be interested in
All accountants at organizations prepare books of accounts in terms of money. Which function of money does this represent? A. me
Masja [62]

Answer:

  B. unit of account

Step-by-step explanation:

Accounts are kept in terms of units of account. Money serves as the unit of account when accounts are kept in terms of money.

6 0
3 years ago
A 20-gallon salt-water solution contains 15% pure salt. How much pure water should be added to it to produce a 10% solution?
Elanso [62]

15% of 20 gal = 0.15 * 20 gal = 3 gal, so the solution contains 3 gal of salt.

If we add <em>x</em> gal of water to the solution, we end up with (20 + <em>x</em>) gal of solution. We want the new mixture to have a concentration of 10%, or

10% of (20 + <em>x</em>) gal = 0.1 * (20 + <em>x</em>) gal = 2 + 0.1<em>x</em> gal

of salt.

The amount of salt in the tank hasn't changed. Solve for <em>x </em>:

2 + 0.1<em>x</em> = 3

0.1<em>x</em> = 1

<em>x</em> = 10 gal

3 0
3 years ago
A person gets on a Ferris wheel at the starting point. The starting point is 15 feet off the ground. The Ferris wheel has a radi
lisov135 [29]
<h2>Answer:</h2>

\boxed{\frac{25}{2}(\sqrt{6}+\sqrt{2})+65}

<h2>Explanation:</h2>

<em>The height of the rider from the ground after the Ferris wheel</em> equals <em>the starting point is 15 feet off the ground </em><em>plus </em><em>the radius of the Ferris wheel </em><em>plus </em><em>the opposite side of the triangle ΔABC </em>(See figure below).<em> </em>Hence our goal is to find this side. We have the angle 11\pi /2=165^{\circ}, therefore the angle ∠BAC = 165° - 90° = 75°. So this side can be found using trigonometry:

\overline{BC}=\overline{AB}sin(75^{\circ})=50sin(75^{\circ})=\frac{25}{2}(\sqrt{6}+\sqrt{2})

Finally, the height is:

H=\frac{25}{2}(\sqrt{6}+\sqrt{2})+15+50 \\ \\ \boxed{H=\frac{25}{2}(\sqrt{6}+\sqrt{2})+65}

6 0
3 years ago
Julia went to the movies and bought one jumbo popcorn and two chocolate chip cookies for $5.00. Marvin went to the same movie an
hichkok12 [17]
X=price of one jumbo popcorn
y=price of one chocolate chip cookies

$5.00=$5.00(100 cts / $)=500 cts
$6.00=$6.00(100 cts / $)=600 cts

We suggest this system of equations:

x+2y=500
x+4y=600

we solve this system of equations by reduction method.

-(x+2y=500)
  x+4y=600
----------------------
      2y=100    ⇒ y=100/2=50

x+2y=500
x+2(50)=500
x+100=500
x=500-100
x=400

solution: one chocolate chip cookie cost 50 cts.
4 0
3 years ago
Remember to show work and explain. Use the math font.
MrMuchimi

Answer:

\large\boxed{1.\ f^{-1}(x)=4\log(x\sqrt[4]2)}\\\\\boxed{2.\ f^{-1}(x)=\log(x^5+5)}\\\\\boxed{3.\ f^{-1}(x)=\sqrt{4^{x-1}}}

Step-by-step explanation:

\log_ab=c\iff a^c=b\\\\n\log_ab=\log_ab^n\\\\a^{\log_ab}=b\\\\\log_aa^n=n\\\\\log_{10}a=\log a\\=============================

1.\\y=\left(\dfrac{5^x}{2}\right)^\frac{1}{4}\\\\\text{Exchange x and y. Solve for y:}\\\\\left(\dfrac{5^y}{2}\right)^\frac{1}{4}=x\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\dfrac{(5^y)^\frac{1}{4}}{2^\frac{1}{4}}=x\qquad\text{multiply both sides by }\ 2^\frac{1}{4}\\\\\left(5^y\right)^\frac{1}{4}=2^\frac{1}{4}x\qquad\text{use}\ (a^n)^m=a^{nm}\\\\5^{\frac{1}{4}y}=2^\frac{1}{4}x\qquad\log_5\ \text{of both sides}

\log_55^{\frac{1}{4}y}=\log_5\left(2^\frac{1}{4}x\right)\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\dfrac{1}{4}y=\log(x\sqrt[4]2)\qquad\text{multiply both sides by 4}\\\\y=4\log(x\sqrt[4]2)

--------------------------\\2.\\y=(10^x-5)^\frac{1}{5}\\\\\text{Exchange x and y. Solve for y:}\\\\(10^y-5)^\frac{1}{5}=x\qquad\text{5 power of both sides}\\\\\bigg[(10^y-5)^\frac{1}{5}\bigg]^5=x^5\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(10^y-5)^{\frac{1}{5}\cdot5}=x^5\\\\10^y-5=x^5\qquad\text{add 5 to both sides}\\\\10^y=x^5+5\qquad\log\ \text{of both sides}\\\\\log10^y=\log(x^5+5)\Rightarrow y=\log(x^5+5)

--------------------------\\3.\\y=\log_4(4x^2)\\\\\text{Exchange x and y. Solve for y:}\\\\\log_4(4y^2)=x\Rightarrow4^{\log_4(4y^2)}=4^x\\\\4y^2=4^x\qquad\text{divide both sides by 4}\\\\y^2=\dfrac{4^x}{4}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\y^2=4^{x-1}\Rightarrow y=\sqrt{4^{x-1}}

6 0
3 years ago
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