80 POINTS!!!!!!

Kipish [7]

Considering that the powers of 7 follow a pattern, it is found that the last two digits of $7^{1867}$ are 43.

<h3>What is the powers of 7 pattern?</h3>

The last two digits of a power of 7 will always follow the following pattern: {07, 49, 43, 01}, which means that, for $7^n$ , we have to look at the remainder of the division by 4:

If the remainder is of 1, the last two digits are 07.

If the remainder is of 2, the last two digits are 49.

If the remainder is of 3, the last two digits are 43.

If the remainder is of 0, the last two digits are 01.

In this problem, we have that n = 1867, and the remainder of the division of 1867 by 4 is of 3, hence the last two digits of $7^{1867}$ are 43.

More can be learned about the powers of 7 pattern at brainly.com/question/10598663

3 0

2 years ago

Choose the constant term that completes the perfect square trinomial.

BigorU [14]

A perfect trinomial, if we begin by a binomial is defined as: the square of the first term, plus (or minus) the double product of the first term times the second, plus the square of the second term:
(a + b)^2 = a^2 + 2ab + b^2
We are given:
y^2 + 5y + x
we need to find x, so x is defined as a squared quantity, which is equal to the second term coefficient (5) divided by 2, and that number squared, that is:
(5/2)^2 = 25/4
that is the third term for the trinomial to be perfect.

3 0

3 years ago

Which statement is true about the function f(x) = -x?

Nataly [62]

Answer:

The answer is "It has the same domain as the function f(x) = --x".

Step-by-step explanation:

If we consider its parent function that is: y= x

Domain function is: $x \in (0, \infty)$

The range function is: $y \in (0, \infty)$

The function has both the same (domain and range).

6 0

3 years ago

Which is the inverse of the function a(d)=5d-3? And use the definition of inverse functions to prove a(d) and a-1(d) are inverse

Drupady [299]

Answer:

$a'(d) = \frac{d}{5} + \frac{3}{5}$

$a(a'(d)) = a'(a(d)) = d$

Step-by-step explanation:

Given

$a(d) = 5d - 3$

Solving (a): Write as inverse function

$a(d) = 5d - 3$

Represent a(d) as y

$y = 5d - 3$

Swap positions of d and y

$d = 5y - 3$

Make y the subject

$5y = d + 3$

$y = \frac{d}{5} + \frac{3}{5}$

Replace y with a'(d)

$a'(d) = \frac{d}{5} + \frac{3}{5}$

Prove that a(d) and a'(d) are inverse functions

$a'(d) = \frac{d}{5} + \frac{3}{5}$ and $a(d) = 5d - 3$

To do this, we prove that:

$a(a'(d)) = a'(a(d)) = d$

Solving for $a(a'(d))$

$a(a'(d)) = a(\frac{d}{5} + \frac{3}{5})$

Substitute $\frac{d}{5} + \frac{3}{5}$ for d in $a(d) = 5d - 3$

$a(a'(d)) = 5(\frac{d}{5} + \frac{3}{5}) - 3$

$a(a'(d)) = \frac{5d}{5} + \frac{15}{5} - 3$

$a(a'(d)) = d + 3 - 3$

$a(a'(d)) = d$

Solving for: $a'(a(d))$

$a'(a(d)) = a'(5d - 3)$

Substitute 5d - 3 for d in $a'(d) = \frac{d}{5} + \frac{3}{5}$

$a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}$

Add fractions

$a'(a(d)) = \frac{5d - 3+3}{5}$

$a'(a(d)) = \frac{5d}{5}$

$a'(a(d)) = d$

Hence:

$a(a'(d)) = a'(a(d)) = d$

7 0

2 years ago

Please help with the question in the picture!

Stella [2.4K]

Answer:

Tan C = 3/4

Step-by-step explanation:

Given-

∠ A = 90°, sin C = 3 / 5

METHOD - I

Sin² C + Cos² C = 1

Cos² C = 1 - Sin² C

Cos² C = $1 - \frac{9}{25}$

Cos² C = $\frac{25 - 9}{25}$

Cos² C = $\frac{16}{25}$

Cos C = $\sqrt{\frac{16}{25} }$

Cos C = $\frac{4}{5}$

As we know that

Tan C = $\frac{Sin C}{Cos C }$

Tan C = $\frac{\frac{3}{5} }{\frac{4}{5} }$

Tan C = $\frac{3}{4}$ 

METHOD - II

Given Sin C = $\frac{3}{5} = \frac{Height}{Hypotenuse}$

therefore,

AB ( Height ) = 3; BC ( Hypotenuse) = 5

∵ ΔABC is Right triangle.

∴ By Pythagorean Theorem-

AB² + AC² = BC²

AC² = BC² - AB²

AC² = 5² - 3²

AC² = 25 - 9

AC² = 16

AC ( Base) = 4

Since,

Tan C = $\frac{Height}{Base}$ 

Tan C = $\frac{AB}{AC}$

Hence Tan C = $\frac{3}{4}$ 

4 0

3 years ago