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3 years ago

HELP PLEASE! 40 POINTS AND WILL MARK BRAINLIEST.

Mathematics

1 answer:

pshichka [43]3 years ago

3 0

Answer:

Explicit rule, because you can find the 40th term directly using the explicit rule. You would need to find the first 39 terms before finding the 40th terms using recursive rule.

hope this helps...

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Find the size of angle YZX

Tasya [4]

Answer:

21.001 cm

Step-by-step explanation:

The diagram shows right triangle with

Hypotenuse ZY = 22.3 cm

Leg XY = 7.5 cm

By the Pythagorean theorem,

$XY^2+XZ^2=ZY^2\\ \\7.5^2+XZ^2=22.3^2\\ \\XZ^2=22.3^2-7.5^2\\ \\XZ^2=(22.3-7.5)(22.3+7.5)\\ \\XZ^2=14.8\cdot 29.8\\ \\XZ^2=441.04\\ \\XZ=\sqrt{441.04}\approx 21.001\ cm$

3 0

3 years ago

Find all points on the x-axis that are 14 units from the point (4, -7).

BaLLatris [955]

Answer:

The points are: (16.12,0),(-8.12,0).

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

Distance between two points:

Suppose we have two points, $(x_1,y_1)$ and $(x_2,y_2)$ . The distance between them is given by:

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Find all points on the x-axis that are 14 units from the point (4, -7).

Being on the x-axis mean that they have y-coordinate equal to 0, so the point is (x,0).

The distance is 14. So

$\sqrt{(x-4)^2+(0-(-7))^2} = 14$

$\sqrt{x^2 - 8x + 16 + 49} = 14$

$\sqrt{x^2 - 8x + 65} = 14$

$(\sqrt{x^2 - 8x + 65})^2 = 14^2$

$x^2 - 8x + 65 - 196 = 0$

$x^2 - 8x - 131 = 0$

So $a = 1, b = -8, c = -131$

$\bigtriangleup = (-8)^{2} - 4(1)(-131) = 588$

$x_{1} = \frac{-(-8) + \sqrt{588}}{2} = 16.12$

$x_{1} = \frac{-(-8) - \sqrt{588}}{2} = -8.12$

The points are: (16.12,0),(-8.12,0).

4 0

2 years ago

For all integers n, if n2 is odd, then n is odd. Use a proof by contraposition, as in Lemma 1.1. Let n be an integer. Suppose th

Reil [10]

Let's assume that the statement "if n^2 is odd, then is odd" is false. That would mean "n^2 is odd" leads to "n is even"

Suppose n is even. That means n = 2k where k is any integer.

Square both sides

n = 2k
n^2 = (2k)^2
n^2 = 4k^2
n^2 = 2*(2k^2)

The expression 2(2k^2) is in the form 2m where m is an integer (m = 2k^2) which shows us that n^2 is also even.

So this contradicts the initial statement which forces n to be odd.

8 0

3 years ago

Which function is nonlinear

viva [34]

the correct answer will be c.

5 0

3 years ago

Read 2 more answers

17<br><br> i already did the test

Dmitry [639]

Answer:

Step-by-step explanation:

I believe the answer is 17.

Simplify

=17

Now,

Consider point O at zero on the number line.

Mark a point A such at number 4 such that OA= 4

Draw a perpendicular AB at point A of unit length.

Join AB

Taking radius of length AB draw an arc to intersect the number line at C.

Finally, OC=\sqrt{17}

Thus proving the answer is 17 or $\sqrt{289} = 17$

3 0

3 years ago