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2 years ago

Can someone help me Plz

Mathematics

1 answer:

Dominik [7]2 years ago

7 0

Answer:

Step-by-step explanation:

Please give brainliest

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In what rate of interest Rs 4000 amount to Rs 6000 in 2year

PIT_PIT [208]

Principal = Rs 4000

Amount = Rs 6000

Time = 2 years

Rate = ?

Amount = principal + simple interest

Rs 6000 = Rs 4000 + simple interest

simple interest = Rs 6000 - Rs 4000 = Rs 2000

Simple interest = P*R*T/100

Rs 2000 = Rs 4000* R*2/100

R = 100/4 = 25%

8 0

2 years ago

Based on Pythagorean identities, which equation is true? A. Sin^2 theta -1= cos^2 theta B. Sec^2 theta-tan^2 theta= -1 C. -cos^2

Arturiano [62]

Answer:

Step-by-step explanation:

our basic Pythagorean identity is cos²(x) + sin²(x) = 1

we can derive the 2 other using the listed above.

1. (cos²(x) + sin²(x))/cos²(x) = 1/cos²(x)

1 + tan²(x) = sec²(x)

2.(cos²(x) + sin²(x))/sin²(x) = 1/sin²(x)

cot²(x) + 1 = csc²(x)

A. sin^2 theta -1= cos^2 theta

this is false

cos²(x) + sin²(x) = 1

isolating cos²(x)

cos²(x) = 1-sin²(x), not equal to sin²(x)-1

B. Sec^2 theta-tan^2 theta= -1

1 + tan²(x) = sec²(x)

sec²(x)-tan(x) = 1, not -1

false

C. -cos^2 theta-1= sin^2

cos²(x) + sin²(x) = 1

sin²(x) = 1-cos²(x), our 1 is positive not negative, so false

D. Cot^2 theta - csc^2 theta=-1

cot²(x) + 1 = csc²(x)

isolating 1

1 = csc²(x) - cot²(x)

multiplying both sides by -1

-1 = cot²(x) - csc²(x)

TRUE

3 0

3 years ago

You invest $5,000 in a savings account that earns 6% interest each year.

weqwewe [10]

A. Multiply $5000 by .06, and add it back to $5000 to get the total after one year. This is $5300.
B. Multiply the new amount, or $5300 by .06, and add it back to $5300 to get the total after 2 years. This is $5618.

A. $5300
B. $5618

4 0

3 years ago

Find the coordinates of the midpoint of the segment with the given endpoints.

uysha [10]

$(\frac{x_{1} + y_{1}}{2}, \frac{x_{2} + y_{2}}{2}) \\(\frac{6 + 10}{2}, \frac{-3 + 5}{2}) \\(\frac{16}{2}, \frac{2}{2}) \\(8, 1)$

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2 years ago

Which expression is equivalent to square root 55x^7 y^6/11x^11 y^8? Assume x > 0 and y > 0.

morpeh [17]

The first step for solving this expression is to reduce the fraction with 11.
$\sqrt{ \frac{5 x^{7} y^{6} }{ x^{11} y^{8} } }$
Now reduce the fraction with $x^{7}$ .
$\sqrt{ \frac{5 y^{6} }{ x^{4} y^{8} } }$
Reduce the fraction one more time with $y^{6}$ .
$\sqrt{ \frac{5}{ x^{4} y^{2} } }$
To take the root of a fraction,, we must take the root of the numerator and denominator separately. This rule will change the expression to the following:
$\frac{ \sqrt{5} }{ x^{2} y}$
Since we cannot simplify this fraction any further,, the final answer is going to be $\frac{ \sqrt{5} }{ x^{2} y}$ .
Let me know if you have any further questions.
:)

3 0

2 years ago

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