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2 years ago

If Triangle DEF is congruent to Triangle ABC and AB=4.4 units, what is the length of DE?

Mathematics

1 answer:

Nana76 [90]2 years ago

6 0

Answer:

8.8

Step-by-step explanation:

If DEF///ABC

AB=4.4

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Answer with yes or no please

KiRa [710]

Answer:lol no

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Find the following:

Butoxors [25]

Answer:

Step-by-step explanation:

Limit refers to the value that the function approaches as the input approaches some value.

We say $\displaystyle \lim_{x\rightarrow a}f(x)=L$ , if f(x) approaches L as x approaches 'a'.

(a)

$\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=4\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=4\\$

$\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-8}{\displaystyle \lim_{x\rightarrow 5}x-5}=4\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4\left ( \displaystyle \lim_{x\rightarrow 5}x-5 \right )\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4\displaystyle \lim_{x\rightarrow 5}x-4(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4(5)-4(5)\\$

$\displaystyle \lim_{x\rightarrow 5}f(x)-8=20-20=0\\\displaystyle \lim_{x\rightarrow 5}f(x)=8$

(b)

$\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-8}{\displaystyle \lim_{x\rightarrow 5}x-5}=7\\$

$\displaystyle \lim_{x\rightarrow 5}f(x)-8=7\left ( \displaystyle \lim_{x\rightarrow 5}x-5 \right )\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=7\displaystyle \lim_{x\rightarrow 5}x-7(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=7(5)-7(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=35-35=0\\\displaystyle \lim_{x\rightarrow 5}f(x)=8$

3 0

4 years ago

What values of c and d make the equation true? Assume x>0 and 20.

Grace [21]

Answer:

c=1 and c=3

Step-by-step explanation:

If x>0 and y>0, then

\sqrt{\dfrac{50x^6y^3}{9x^8}}=\sqrt{\dfrac{25\cdot 2y^2\cdot y}{9x^2}}=\dfrac{5y\sqrt{2y}}{3x}.

50x

25⋅2y

⋅y

\dfrac{5y\sqrt{2y}}{3x}

is equal to

\dfrac{5y^c\sqrt{2y}}{dx},

then

\begin{gathered}y=y^c\Rightarrow c=1,\\ \\3x=dx\Rightarrow d=3.\end{gathered}

y=y

⇒c=1,

3x=dx⇒d=3.

5 0

2 years ago

Kittens weigh about 100 grams when born and gain 7 to 15 grams per day. If a kitten weighed 100 grams at birth and gained 8 gram

tangare [24]

25 days

First, start with an equation

100+8x= kittens weight

(X representing the days)

Since you’re trying to figure out the number of days it takes the kitten to triple it’s weigh you multiply it’s initial weight by 3 (100•3)

Then, you output 300 into the equation and solve

8 0

4 years ago

The radioactive substance cesium-137 has a half-life of 30 years. The amount A (t)(in grams) of a sample of cesium-137 remaining

lakkis [162]

Rather than use that formula, look at the attachment and use the 4th equation.

Ending Amount = Beginning Amount / 2^n
where "n" is the number of half-lives. "n" = elapsed time / half-life
(Is 934 the beginning amount? Let's say it is.)

for 40 years, n=40/30 = 1.3333333...
Ending Amount = 934 / 2^1.3333333...
Ending Amount = 934 / 2.5198420998 
Ending Amount = 370.66 grams

for 100 years, n=100/30 = 3.333333 ...
Ending Amount = 934 / 2^ 3.333333 ...
Ending Amount = 934 / 10.0793683992 
Ending Amount = 92.66 grams

Source:
http://www.1728.org/halflife.htm

4 0

3 years ago