Please help asap!!! thank ya

Mathematics

1 answer:

enot [183]2 years ago

3 0

22.7 units
Hope that helps!

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In the diagram below, GH is parallel to DE . If GH is 3 more than EH FH=10, and DE=12 find the length of EH. Figures are not nec

steposvetlana [31]

Answer: 5

Step-by-step explanation:

Since $\overline{GH} \parallel \overline{DE}$ , by the corresponding angles theorem, $\angle FGH \cong \angle FDE, \angle FHG \cong \angle FED$ . This means $\triangle FGH \sim \triangle FDE$ by AA.

As corresponding sides of similar triangles are proportional,

$\frac{10}{x+3}=\frac{10+x}{12}\\(10)(12)=(10+x)(x+3)\\120=x^{2}+13x+30\\x^{2}+13x-90=0\\(x+18)(x-5)=0\\x=-18, 5$
However, as distance must be positive, we consider the positive solution, x=5.

Therefore, the answer is <u>5</u>

6 0

2 years ago

Whats the answer to 3,000(6-2) + 3,000

yuradex [85]

The answer is 15000

$3000 \times ( 6 - 2) = 12000 + 3000 = 15000$
3000*6=1800
3000*-2=-6000
18000-6000=12000
12000+3000=15000

8 0

3 years ago

PLEASE HELP!!!

Scilla [17]

You need the Law of Cosines here and you use it when you have 2 sides and an enclosed angle. The side across from the angle is the one we are looking for. The correct way to express the Law using what we have is the last choice above. Side RT is the unknown, and it is across from the angle that is enclosed between the 2 other sides.

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3 years ago

Round 6700 to the nearest thousand

andriy [413]

Answer:

Step-by-step explanation:

it would be 7000 because the 7 is greater than five

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3 years ago

Airen’s grandparents deposited $1300 in a mutual fund earning 6% interest compounded annually. Write an equation to represent ho

IRINA_888 [86]

$~~~~~~ \textit{Compound Interest Earned Amount \underline{in 18 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$1300\\ r=rate\to 6\%\to \frac{6}{100}\dotfill &0.06\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &18 \end{cases}$

$A=1300\left(1+\frac{0.06}{1}\right)^{1\cdot 18}\implies A=1300(1.06)^{18}\implies A\approx 3710.64 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y = 1300(1.06)^x~\hfill$

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2 years ago