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3 years ago

8

What is the solution to -3(x + 4) = -6x + 18

1 answer:

Leno4ka [110]3 years ago

8 0

Answer:

x=10

Step-by-step explanation:

easy

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In order to avoid staying after work, a quality control analyst inspects the first 100 items produced that day. Which sampling m

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Based on the given scenario above wherein the quality control analyst just inspects the first 100 items produced that day in order to avoid staying after work, the sampling method that she used is the CONVENIENCE SAMPLING METHOD. Hope this helps.

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Given: ABC is a right triangle with right angle C. AC=15 centimeters and m∠A=40∘ . What is BC ? Enter your answer, rounded to th

konstantin123 [22]

In order to answer this question, the figure in the first picture will be helpful to understand what a right triangle is. Here, a right angle refers to $90\°$ .

However, if we want to solve the problem we have to know certain things before:

In the second figure is shown a general right triangle with its three sides and another given angle, we will name it $\alpha$ :

The side <u>opposite to the right angle</u> is called The Hypotenuse (h)

The side <u>opposite to the angle $\alpha$ </u> is called the Opposite (O)

The side <u>next to the angle $\alpha$ </u> is called the Adjacent (A)

So, going back to the triangle of our question (first figure):

The Hypotenuse is AB

The Opposite is BC

The Adjacent is AC

Now, if we want to find the length of each side of a right triangle, we have to use the <u>Pythagorean Theorem</u> and T<u>rigonometric Functions:</u>

Pythagorean Theorem

$h^{2}=A^{2} +O^{2}$ (1)

Trigonometric Functions (here are shown three of them):

Sine: $sin(\alpha)=\frac{O}{h}$ (2)

Cosine: $cos(\alpha)=\frac{A}{h}$ (3)

Tangent: $tan(\alpha)=\frac{O}{A}$ (4)

In this case the function that works for this problem is cosine (3), let’s apply it here:

$cos(40\°)=\frac{AC}{h}$

$cos(40\°)=\frac{15}{h}$ (5)

And we will use the Pythagorean Theorem to find the hypotenuse, as well:

$h^{2}=AC^{2}+BC^{2}$

$h^{2}=15^{2}+BC^{2}$ (6)

$h=\sqrt{225+BC^2}$ (7)

Substitute (7) in (5):

$cos(40\°)=\frac{15}{\sqrt{225+BC^2}}$

Then clear BC, which is the side we want:

${\sqrt{225+BC^2}}=\frac{15}{cos(40\°)}$

${{\sqrt{225+BC^2}}^2={(\frac{15}{cos(40\°)})}^2$

$225+BC^{2}=\frac{225}{{(cos(40\°))}^2}$

$BC^2=\frac{225}{{(cos(40\°))}^2}-225$

$BC=\sqrt{158,41}$

$BC=12.58$

Finally $BC$ is approximately 13 cm

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What kind of triangle is this?

NeTakaya

Answer:

Obtuse

Step-by-step explanation:

Has one obtuse angle, i hope this helps you

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