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3 years ago

Solve absolute value equation. Check for extraneous solution. |2x-3|=2x-1. Post it into the drop box for the unit 2 review.

1 answer:

6 0

let's keep in mind that, an absolute value expression is in effect a piecewise one, since it has a ± versions of it.

$\bf |2x-3|=2x-1\implies \begin{cases}+(2x-3)=2x-1\\-(2x-3)=2x-1\end{cases}\\\\[-0.35em]~\dotfill\\\\+(2x-3)=2x-1\implies 2x-3=2x-1\implies -3 \ne -1\impliedby extraneous\\\\[-0.35em]~\dotfill\\\\-(2x-3)=2x-1\implies 2x-3=-2x+1\implies 4x=4\\\\\\x=\cfrac{4}{4}\implies x=1$

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Multiply 42 by the sum of 4 and r. I need help on number 10 please.

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Answer:

42(4 + r) is the answer if I'm reading that correctly.

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3 years ago

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3 years ago

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What is the length of the altitude of an equilateral triangle with side a?

nikitadnepr [17]

Answer:

(a√3)/2

Step-by-step explanation:

You can find the height (h) of a equilateral triangle any of several ways. A simple one is to use the Pythagorean theorem.

In the diagram below, h is one leg of the right triangle with a/2 as the other leg and a as the hypotenuse. The Pythagorean theorem gives the relationship as ...

h² + (a/2)² = a²

h² = a² - a²/4 . . . . . . . subtract (a/2)²

h² = a²(1 -1/4) = a²(3/4)

h = √(a²(3/4)) = (a/2)√3

The height is (√3)/2 times the side length.

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3 years ago

ice cream delight uses 500 milliliters of milk in each milkshake. if the shop sells 4 milkshakes, how many liters of milk has be

Elina [12.6K]

500 milliliters x 4 = 2000
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3 years ago

WILL GIVE BRAINIEST IF DONE IN 10 MINUTES!!!!!

Mama L [17]

Answer:

$\sin(\theta)=-12/13\text{ and } \csc(\theta)=-13/12\\\cos(\theta)=-5/13\text{ and } \sec(\theta)=-13/5\\\tan(\theta)=12/5\text{ and } \cot(\theta)=5/12$

Step-by-step explanation:

So we know that:

$\sin(\theta)=-12/13\text{ and } \sec(\theta)$

This tells us that our angle θ is in Quadrant III. This is because, recall ASTC. In QI, everything is positive. This is not the case here since sine and secant are both negative.

In QII, only sine is positive. Since sine is negative, θ can't be in QII.

In QIII, only tangent (and cotangent) is positive. All other are negative. So, θ is in QIII.

And just to check, in QIV, only cosine (and secant) is positive. Since we are told that secant is less than 0 (in other words, negative), θ can't be in QIV.

Now that we know that θ is in QIII, we know that sine and cosecant is negative; cosine and secant is negative; and tangent and cotangent is positive.

Now, let's find the ratios. Remember what sine tells us. Sine gives us the ratio of the opposite side to the hypotenuse. With this information, let's find the adjacent side:

So, the opposite is 12 (ignore the negative for now) and the hypotenuse is 13. Therefore, we can use the Pythagorean Theorem:

$a^2+b^2=c^2$

Substitute 12 for a and 13 for c:

$12^2+b^2=c^2$

Now, solve for b to find the adjacent side. Square the values:

$144+b^2=169$

Subtract 144 from both sides:

$b^2=25$

Square root:

$b=\pm 5$

We can ignore the negative and just take the positive. While the answer is technically -5, we don't need to do that. So, our adjacent side is 5.

Therefore, with respect to angle θ, our opposite is 12, our adjacent is 5, and our hypotenuse is 13. With this, we can compute the other trig ratios. Let's start with the 3 main ones:

Sine:

$\sin(\theta)$