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liraira [26]
3 years ago
9

G evaluate the given integral by changing to polar coordinates. e−x2 − y2 da d where d is the region that lies to the left of th

e y-axis between the circles x2 + y2 = 1 and x2 + y2 = 25.
Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
7 0
Wages wages wages wages wages wages wages wages wages wages wages wages wages 
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Can someone help me with this please
MaRussiya [10]
X,y is 5(5y,-2y) and 5(y-2)
3 0
3 years ago
A wedding planner uses 72 ivy stems for 18 centerpieces when she arrives at the venue she realizes she will only need 16 centerp
beks73 [17]

Answer:

64 ivy stems

Step-by-step explanation:

To solve this problem we can use rule of three:

Firstly we had 72 ivy stems to use in 18 centerpieces, and know we have 16 centerpieces and want to know how many ivy stems we need to use, so:

18 centerpieces -> 72 ivy stems

16 centerpieces -> X ivy stems

18/16 = 72/X

X = 16*72/18 = 64

He needs to use now 64 ivy stems to maintain the ratio.

6 0
3 years ago
Read 2 more answers
Can somebody please help me???
matrenka [14]

Answer: B

Step-by-step explanation:

Domain is x values

Range is y values

Domain you look from left to right. You can see it starts at -3, and it goes all the way to 2. NOT 4 BUT 2!! Since we are looking at the X.

Range.. WE ARE LOOKING AT THE LOWEST POINT!! LOWEST POINT FIRST!1 I cannot stress enough. The lowest y-value is -2, and it's maximum point is 4!

REMEMBER:

DOMAIN = X LEFT TO RIGHT

RANGE= Y LOWEST POINT TO HIGHEST

8 0
3 years ago
Read 2 more answers
Nemos aquarium is filled with 2400 cm of water the base of the aquarium is 20 cm long and 12 cm wide what is the height of the w
SSSSS [86.1K]

The answer to this question is 10 cm

In this question, you asked the height and given the volume of water(2400cm3), aquarium length(20cm), and aquarium width(12cm).

Then you need to divide the volume with the length and width. The equation would be:

Aquarium height= volume / length / width

Aquarium height= 2400 cm3 / 20cm / 12 cm = 10cm.

7 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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