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miskamm [114]
3 years ago
5

You may remember that the area of a triangle is A=12B⋅H, where B is the base of the triangle and H is the height.

Mathematics
2 answers:
Alexxandr [17]3 years ago
7 0

Answer:

b = 6

Step-by-step explanation:

The area of a triangle is

A = 1/2 bh

9 = 1/2 *(2h) * h

9 = h^2

Take the square root of each side

3 =h

The base is 2h

b = 2(3)

b = 6

notsponge [240]3 years ago
3 0

Answer:

6

Step-by-step explanation:

Assuming that the base is b and the height is h:

\dfrac{bh}{2}=9 \\\\b=2h

Substitute:

\dfrac{2h(h)}{2}=9 \\\\h^2=9 \\\\h=3 \\\\b=2h=6

Hope this helps!

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A

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A shop keeper has offered an item for sale. Its label price is Rs . It is not sold in one-month period and after one month its l
BigorU [14]

Answer:

x = Rs 20,833.33

the value of x is Rs 20,833.33

Step-by-step explanation:

Let x,y and z represent the price of the item initially, after one month and after two months respectively.

Given that;

after one month its label price is reduced by 20%

y = x - 20% of x

y = x - 0.20x

y = 0.80x ........1

after 2 months its reduced price is further reduced by 10% and then sold it for Rs 15000.

z = y - 10% of y

z = y - 0.10y

z = 0.90y ........2

Substituting equation 1 into 2;

z = 0.90(0.80x)

z = 0.72x

Also z = Rs 15000

So,

z = 0.72x = Rs 15000

0.72x = Rs 15000

x = Rs 15000/0.72

x = Rs 20833.33333333

x = Rs 20,833.33

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5 0
3 years ago
Solve for n: 2n + 3 + 3n = n + 11
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N is equal to 2 in the equasion

4 0
3 years ago
Read 2 more answers
!! AP CALCULUS AB !!<br> find the limit of [(secx*sinx) + (cscx*cosx)]/(3secx) as x approaches 3pi/2
uysha [10]

f(x)= \frac{\frac{\sin{x}}{\cos{x}}+\frac{\cos{x}}{\sin{x}}}{3\sec{x}}

\frac{\frac{\sin{x}}{\cos{x}}+\frac{\cos{x}}{\sin{x}}}{3\sec{x}} \implies \\ \frac{\sin^2{x}+\cos^2{x}}{\sin{x}\cos{x}}\frac{\cos{x}}{3}\implies \\\frac{1}{\sin{x}\cos{x}}*\frac{\cos{x}}{3}\implies \\ \frac{1}{3\sin{x}}

So,

\lim_{x \rightarrow \frac{3\pi}{2}}{f(x)=f(\frac{3\pi}{2})=\frac{1}{3\sin{\frac{3\pi}{2}}}=-\frac{1}{3}

6 0
3 years ago
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