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3 years ago

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Determine if the statement is true or false. A linear system with three equations and two variables must be inconsistent. True F

alse Justify your answ

1 answer:

Arturiano [62]3 years ago

6 0

Answer:

yes it is true that a linear system of three equations with two variables is inconsistent because when we solve the equations variables can be obtained by only first two equations which are considered first. and third equation is useless.

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THE VOLUME OF A SPHERE IS 38808cm^3. find the height of the cone

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I am not 100% positive but I think it is 4117.66

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Find the volume of the following compound shape.<br>

Bess [88]

<h3>Given:</h3>

Cone
Cylinder

<h3>Volume of the cone:</h3>

$v = \frac{1}{3} \pi {r}^{2} h$

$v = \frac{1}{3} \times \pi \times {10}^{2} \times 10$

$v = 1047.20 \: {cm}^{3}$

<h3>Volume of the cylinder:</h3>

$v = \pi {r}^{2} h$

$v = \pi \times {10}^{2} \times 10$

$v = 3141.59 \: {cm}^{3}$

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2 years ago

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In the following diagram, line m is parallel to line n and line q intersects the parallel lines at

joja [24]

Answer:

108°

Step-by-step explanation:

We know for a theorem that A = B, so:

4x + 28 = 2x + 68

Isolating the x:

2x = 40

x = 20

Now let's plug in the equation for A:

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2 years ago

True or False? A circle could be circumscribed about the quadrilateral below.

KengaRu [80]

The answer is false your welcome

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3 years ago

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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:

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

$r = \sqrt{50$

$r = \sqrt{25 * 2$

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

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

<u>Solving the trigonometry functions</u>

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

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

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