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3 years ago

Please help me solve the problem

Mathematics

1 answer:

Elina [12.6K]3 years ago

7 0

Answer:

x = $\sqrt{2}$ ; y = 2

Step-by-step explanation:

You have a 45 -45 - 90 triangle. There are two things to know"

1. The hypothenuse = leg × $\sqrt{2}$

2. The legs are equal.

The hypothenuse is opposite the right angle that is "y" in this problem.

Hypothenuse = leg × $\sqrt{2}$

= $\sqrt{2} \sqrt{2} = \sqrt{4} = 2$

The legs are =

Therefore x = $\sqrt{2}$

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The first triangle is an enlargement of the second.

insens350 [35]

Answer:

large triangle:

base = 6.64 cm

hypotenuse = 8.31 cm

small triangle:

base = 4 cm

Step-by-step explanation:

large triangle:

tan 37 = 5/base

0.7536 = 5/base

base = 6.64 cm

using the Pythagorean theorem:

hypotenuse² = 5² + 6.64²

hypotenuse² = 25 = 44.02 = 69.02

hypotenuse = 8.31 cm

small triangle:

using the Pythagorean theorem:

base² = 5² - 3²

base² = 25 - 9 = 16

base = 4 cm

3 0

3 years ago

What must be true of PQ?

yulyashka [42]

Interesting question
Usually when you look at something like that construction, you think that AB has been bisected by PQ and that the two segments are perpendicular. They are perpendicular but nowhere is that stated. So the answer is C because all the other answers are wrong.

PQ is congruent AB is not correct. As long as the arcs are equal and meet above and below AB there is no proof of congruency. In your mind widen the compass legs so that they are wider than AB and redraw the arcs. You get a larger PQ, but it has all the original properties of PQ except size.

PQ is not congruent to AQ. How would you prove conguency? You'd have to put both lines into triangles that can be proved congruent. It can't be done.

The two lines are not parallel. They are perpendicular. That can be proven. They meet at right angles to each other (also provable).

8 0

3 years ago

Explain why a positive times a negative is a negative number.

prohojiy [21]

Explanation:

This can be explained by thinking numbers on the number line as:

Lets take we have to multiply a positive number (say, 2) with a negative number say (-3)

2×(-3)

Suppose someone is standing at 0 on the number line and to go to cover -3 , the person moves 3 units in the left hand side. Since, we have to compute for 2×(-3), The person has to cover the same distance twice. At last, he will be standing at -6, which is a negative number.

A image is shown below to represent the same.

Thus, a positive times a negative is a negative number.

8 0

4 years ago

Please someone help me solve this. in graphic method.

fiasKO [112]

Answer:

just graph it mate

Step-by-step explanation:

5 0

4 years ago

i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = -5\sqrt 2$

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

$r^2 = x^2 + y^2$

$r = \sqrt{x^2 + y^2$

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

$r = \sqrt{(-7)^2 + (-1)^2$

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

Solving the trigonometry functions

$sin\ \varnothing = \frac{y}{r}$

$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

Rationalize:

$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{x}{r}$

$cos\ \varnothing = \frac{-7}{5\sqrt 2}$

Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

3 0

3 years ago