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Vlad [161]
3 years ago
11

It is given that in △ABC, AD ⊥ BC. Using the definition of sine with angle C in △ACD results in sin(C) = . Using the multiplicat

ion property of equality to isolate h, the equation becomes bsin(C) = h. Knowing that the formula for the area of a triangle is A = bh is and using the side lengths as shown in the diagram, which expression represents the area of △ABC? bsin(C) absin(C) cbsin(C) hbsin(C)

Mathematics
2 answers:
amid [387]3 years ago
7 0

We see here in the diagram that the base is a. We know this because the height is perpendicular to it. We also know the height is bsin(C) which, when replace h for bsin(C) and a for the base, we get A=absin(C), which is the second option.

shtirl [24]3 years ago
6 0

Answer:

B

Step-by-step explanation:

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Can 3.65909090909 be expressed as a fraction whose denominator is a power of 10? Explain.
GuDViN [60]
\bf 3.659\textit{ can also be written as }\cfrac{3659}{1000}\textit{ therefore }3.6590909\overline{09}\\\\
\textit{can be written as }\cfrac{3659.0909\overline{09}}{1000}

notice above, all we did, was isolate the "recurring part" to the right of the decimal point, so the repeating 09, ended up on the right of it.

now, let's say, "x" is a variable whose value is the recurring part, therefore then

\bf \cfrac{3659.0909\overline{09}}{1000}\qquad \boxed{x=0.0909\overline{09}} \qquad \cfrac{3659+0.0909\overline{09}}{1000}\implies \cfrac{3659+x}{1000}

now, the idea behind the recurring part is that, we then, once we have it all to the right of the dot, we multiply it by some power of 10, so that it moves it "once" to the left of it, well, the recurring part is 09, is two digits, so let's multiply it by 100 then, 

\bf \begin{array}{llllllll}
100x&=&09.0909\overline{09}\\
&&9+0.0909\overline{09}\\
&&9+x
\end{array}\quad \implies 100x=9+x\implies 99x=9
\\\\\\
x=\cfrac{9}{99}\implies \boxed{x=\cfrac{1}{11}}\\\\
-------------------------------\\\\
\cfrac{3659.0909\overline{09}}{1000}\qquad \boxed{x=0.0909\overline{09}} \quad \cfrac{3659+0.0909\overline{09}}{1000}\implies \cfrac{3659+x}{1000}
\\\\\\
\cfrac{3659+\frac{1}{11}}{1000}

and you can check that in your calculator.
8 0
3 years ago
Show work please I need help asap
rosijanka [135]

Answer:

12. -11

13. 2

Step-by-step explanation:

12. First recall the order of operations (BEDMAS): Brackets, Exponents, Division, Multiplication, Addition, Subtraction

For question 12 there is a division and its priority is before other operations in the equation so you must divide 16/-2 first which gives you -8.

From there because addition and subtraction is on the same level of order, you would do the question straight as it is shown (but replace 16/-2 with -8.

Your new equation is -10-8+7 = -11

13. follow the same as above using BEDMAS. It may help to look at it with additional brackets: ((-68)/(-4)) + ((5 x (-3))

  • keep in mind there’s an addition between the two

Division and multiplication is on the same level in BEDMAS so first divide -68/-4 = 17

Second multiply 5 x -3 = -15

Now you can combine the two answers using addition: 17 + (-15) = 2

5 0
3 years ago
6-2x=3+5x <br> Solve for x
evablogger [386]

Answer:

x=3/7

Step-by-step explanation:

6-2x=3+5x

subtract 5x from both sides

6-2x-5x=3

subtract 6 from both sides

-2x-5x=3-6

add like terms

-7x=-3

divide both sides by -7

x=3/7

3 0
3 years ago
Read 2 more answers
On her way to work each morning sophia purchase a small cup of coffee for 4.25 from the coffee shop
steposvetlana [31]
What do you want us to answer?
5 0
3 years ago
Read 2 more answers
A father wishes to give his son P200, 000 ten years from now. What amount should he invest if it will earn interest at 10% compo
Archy [21]

Answer:

C. P69,256.82

Step-by-step explanation:

We know that,

The amount formula in compound interest is,

A=P(1+\frac{r_1}{n_1})^{n_1t_1} (1+\frac{r_2}{n_2})^{n_2t_2}.......

Where, P is the principal amount,

r_1, r_2.... are the annual rate for the different periods,

t_1, t_2,..... are the number of year for different periods,

n_1, n_2, n_3... are the number of periods,

Given,

A = P 200,000,

r_1=10%=0.1, n_1=4, t_1=5,r_2=12%=0.12, n_2=1, t_2=5

Thus, by the above formula the final amount would be,

200000=P(1+\frac{0.1}{4})^{4\times 5}(1+\frac{0.12}{1})^{1\times 5}

200000=P(1+0.025)^{20}(1+0.12)^5

200000=P(1.025)^{20}(1.12)^5

\implies P=69,256.824\approx 69,256.82

Option C is correct.

8 0
3 years ago
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