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scZoUnD [109]
3 years ago
15

HEEEELPPPPPPPP!!!!!!!!!!!!!

Mathematics
1 answer:
MAVERICK [17]3 years ago
7 0

Answer:

A

Step-by-step explanation:

You might be interested in
In a 30°-60°-90° triangle, what is the length of the longer leg when the length of the hypotenuse is 18 inches?
max2010maxim [7]

Answer:

9sqrt(3) in.

Step-by-step explanation:

The ratio of the lengths of the sides of a 30-60-90 triangle is:

short leg : long leg : hypotenuse

      1        :   sqrt(3)  :          2

If the short leg measures 1, then the hypotenuse measures 2.

The length of the hypotenuse is twice the length of the short leg.

The length of the long leg is sqrt(3) times the length of the short leg.

If the hypotenuse measures 18 in., then the short leg measures 9 in.

Then the long leg measures 9sqrt(3) in.

8 0
3 years ago
What is the value of p if the line that passes through (2,-4) and (p,8) has a slope of 1/2? ​
Irina18 [472]

Answer:

Step-by-step explanation:

(2,-4)....x1= 2 and y1 = -4

(p,8).....x2 = p and y2 = 8

slope(m) = 1/2

now use the slope formula (y2 - y1) / (x2 - x1) and sub in what we know...

slope(m) = (y2 - y1) / (x2- x1)

1 / 2 = (8 - (-4) / (p - 2)

1 / 2 = (8 + 4) / (p - 2)

1/2 = 12 / (p - 2) .....now multiply both sides by (p - 2)

1/2(p - 2) = 12

1/2p - 1 = 12

1/2p = 12 + 1

1/2p = 13

p = 13 / (1/2)

p = 13 * 2/1

p = 26

so the value of p is 26

6 0
3 years ago
2(5x+7)=16 <br> what the answer?
Marianna [84]

Answer:

x = 1/5

Step-by-step explanation:

2(5x + 7) = 16

2 * 5x + 2 * 7 = 16

10x + 14 = 16

10x = 16 - 14

10x = 2

x = 2/10

x = 1/5

6 0
3 years ago
Use the compound interest formula to determine the final value of the following amount. $1900 at 10.4% compounded monthly for 4.
nikklg [1K]

Answer:

$3027.80

Explanation:

The compound interest formula is the following.

A=P(1+\frac{r}{n})^{nt}

where

A = final amount

P = principle amount

r = interest rate / 100

n = number of compounds per interval

t = time interval

Now in our case,

A = unknown

P = $1900

r = 10.4/100

n = 12 months / year ( because the interest is compounded monthly)

t = 4.5 yrs

Therefore, the compound interest formula gives

A=1900(1+\frac{10.4/100}{12})^{12*4.5}

Using a calculator, we evaluate the above to get

\boxed{A=\$3027.80}

which is our answer!

8 0
1 year ago
<img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csf%5Clim_%7Bx%20%5Cto%200%20%7D%20%5Cfrac%7B1%20-%20%5Cprod%20%5Climits_%
xxTIMURxx [149]

To demonstrate a method for computing the limit itself, let's pick a small value of n. If n = 3, then our limit is

\displaystyle \lim_{x \to 0 } \frac{1 - \prod \limits_{k = 2}^{3} \sqrt[k]{\cos(kx)} }{ {x}^{2} }

Let a = 1 and b the cosine product, and write them as

\dfrac{a - b}{x^2}

with

b = \sqrt{\cos(2x)} \sqrt[3]{\cos(3x)} = \sqrt[6]{\cos^3(2x)} \sqrt[6]{\cos^2(3x)} = \left(\cos^3(2x) \cos^2(3x)\right)^{\frac16}

Now we use the identity

a^n-b^n = (a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)

to rationalize the numerator. This gives

\displaystyle \frac{a^6-b^6}{x^2 \left(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5\right)}

As x approaches 0, both a and b approach 1, so the polynomial in a and b in the denominator approaches 6, and our original limit reduces to

\displaystyle \frac16 \lim_{x\to0} \frac{1-\cos^3(2x)\cos^2(3x)}{x^2}

For the remaining limit, use the Taylor expansion for cos(x) :

\cos(x) = 1 - \dfrac{x^2}2 + \mathcal{O}(x^4)

where \mathcal{O}(x^4) essentially means that all the other terms in the expansion grow as quickly as or faster than x⁴; in other words, the expansion behaves asymptotically like x⁴. As x approaches 0, all these terms go to 0 as well.

Then

\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 2x^2\right)^3 \left(1 - \frac{9x^2}2\right)^2

\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 6x^2 + 12x^4 - 8x^6\right) \left(1 - 9x^2 + \frac{81x^4}4\right)

\displaystyle \cos^3(2x) \cos^2(3x) = 1 - 15x^2 + \mathcal{O}(x^4)

so in our limit, the constant terms cancel, and the asymptotic terms go to 0, and we end up with

\displaystyle \frac16 \lim_{x\to0} \frac{15x^2}{x^2} = \frac{15}6 = \frac52

Unfortunately, this doesn't agree with the limit we want, so n ≠ 3. But you can try applying this method for larger n, or computing a more general result.

Edit: some scratch work suggests the limit is 10 for n = 6.

6 0
2 years ago
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