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3 years ago

Find the measure of angle U

Mathematics

1 answer:

Rudiy273 years ago

5 0

Answer:

Step-by-step explanation:

180-110=70

70+80=150

180-150=30

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Suppose that the perimeter a rectangle is 54 feet, and the length is 13 feet more than the width. Find the width ractangle, in f

Vesnalui [34]

Answer:7feet

Step-by-step explanation:

perimeter(p)=54ft

Width(w) +13=Length(L)

Perimeter=2xLength+2xwidth

P=2xL+2xw

L=w+13

54=2(w+13)+2w

54=2w+26+2w

Collect like terms

54-26=2w+2w

28=4w

Divide both sides by 4

28/4=4w/4

7=w

Width =7feet

8 0

4 years ago

Please help me with this

Sonja [21]

The answer to your question is D) √5x/5x.

I hoped this helped!

6 0

4 years ago

How do I solve these problems? ln(x) = 5.6 + ln(7.5) and log(x) = 5.6 - log(7.5)

SOVA2 [1]

Use the rules of logarithms and the rules of exponents.

... ln(ab) = ln(a) + ln(b)

... e^ln(a) = a

... (a^b)·(a^c) = a^(b+c)

_____

1) Use the second rule and take the antilog.

... e^ln(x) = x = e^(5.6 + ln(7.5))

... x = (e^5.6)·(e^ln(7.5)) . . . . . . use the rule of exponents

... x = 7.5·e^5.6 . . . . . . . . . . . . use the second rule of logarithms

... x ≈ 2028.2 . . . . . . . . . . . . . use your calculator (could do this after the 1st step)

2) Similar to the previous problem, except base-10 logs are involved.

... x = 10^(5.6 -log(7.5)) . . . . . take the antilog. Could evaluate now.

... = (1/7.5)·10^5.6 . . . . . . . . . . of course, 10^(-log(7.5)) = 7.5^-1 = 1/7.5

... x ≈ 53,080.96

8 0

4 years ago

Please help

Elena-2011 [213]

Answer:

Step-by-step explanation:

The distance formula is:

$d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2$

-4 - 2 = -6 squared = 36

4 - -3 = 7 squared = 49

36 + 49 = 85

$\sqrt{85} =$ about 9.2

I'm always happy to help :)

4 0

3 years ago

You are saving money to buy an electric guitar. You deposit $1000 in an account that earns interest compounded annually. The exp

Katena32 [7]

Let's move like a crab, backwards some.

after 2 years?

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$1000\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &2
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 2}\implies A=1000(1.03)^2$

after 3 years?

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$1000\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &3
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 3}\implies A=1000(1.03)^3$

is that enough to pay the $1100?

now, let's write 1000(1+r)² in standard form

1000( 1² + 2r + r²)

1000(1 + 2r + r²)

1000 + 2000r + 1000r²

1000r² + 2000r + 1000 <---- standard form.

8 0

3 years ago