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3 years ago

Which translation will change figure ABCD to figure A'B'C'D'?

Mathematics

1 answer:

Liono4ka [1.6K]3 years ago

8 0

Answer:

5 units left and 7 units up

Step-by-step explanation:

idk i amn as confused as you

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An archaeologist at a dig sets up a coordinate system using string. Two similar artifacts are found one at position (1, 4) and t

myrzilka [38]

Given:

Positions of two artifacts are at points (1, 4) and (5, 2).

To find:

The distance between these two artifacts.

Solution:

Distance formula: The distance between two points is

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula, the distance between two points (1, 4) and (5, 2) is

$d=\sqrt{(5-1)^2+(2-4)^2}$

$d=\sqrt{(4)^2+(-2)^2}$

$d=\sqrt{16+4}$

$d=\sqrt{20}$

$d=4.4721359$

Round to the nearest tenth of a unit.

$d\approx 4.5$

Therefore, the distance between two artifacts is 4.5 units.

4 0

2 years ago

Which number is the greatest? 0.35, 0.2, 0.56, 0.8, 0.09

ludmilkaskok [199]

I do believe it is 56

8 0

2 years ago

Read 2 more answers

Exhibit 6-2 the weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 poun

Evgen [1.6K]

The question is asking for the lower bound of the 95% two tailed Confidence interval of the normally distributed population.

95% C.I. is given by 200 + or - 1.96(25) = 200 + or - 49 = (151, 249)
Therefore, the minimum weight of the middle 95% of players is 151 pounds.

5 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:

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

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

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

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

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

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3 years ago

a report stated that 281,305,769 people visited us national parks in 2010. the author of a travel guide rounded the number of vi

Varvara68 [4.7K]

The author rounded to the hundred thousands place.

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3 years ago