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rosijanka [135]

3 years ago

14

Triangles PQR and SQT are similar triangles.

1 answer:

aleksklad [387]3 years ago

8 0

Answer:

The side lengths of similar triangles are proportional

Step-by-step explanation:

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-4x+3x<4 what is the awenswer to this equation

AlladinOne [14]

Answer:

X > -4

Step-by-step explanation:

Collect the like terms which literally leaves you with -x<4

Then just changes the signs, hence the answer.

Please mark brainliest!

6 0

2 years ago

Convert to standard form y=2(x+3)(x-6)

Nataly_w [17]

Answer:

the answer is 2x²-6x-36

3 0

2 years ago

What are the points for f(x)=3^(x-1)-2

Savatey [412]

Answer: (0, -1 $\frac{2}{3}$ ), (1, -1), (2, 1)

<u>Step-by-step explanation:</u>

Set the exponent equal to zero - <em>that is your anchor point</em>.

Then choose an x-value less than and greater than the anchor point.

$\left\begin{array}{c|c|l}&\underline{\quad x\quad }&\underline{\qquad \qquad f(x)\qquad \qquad \qquad \quad }\\\text{less than anchor}&0&3^{0-1}-2=3^{-1}-2=\frac{1}{3}-2 = -1\frac{2}{3}}\\\\anchor\ point&1&3^{1-1}-2=3^{0}-2=1-2 = -1}\\\\\text{greater than anchor}&2&3^{2-1}-2=3^{1}-2=3-2 = 1}\end{array}\right$

4 0

3 years ago

In Exercises 5-7, find all the exact t-values for which the given statement is true,

bearhunter [10]

Answer: See Below

<u>Step-by-step explanation:</u>

NOTE: You need the Unit Circle to answer these (attached)

5) cos (t) = 1

Where on the Unit Circle does cos = 1?

Answer: at 0π (0°) and all rotations of 2π (360°)

In radians: t = 0π + 2πn

In degrees: t = 0° + 360n

******************************************************************************

$6)\quad sin (t) = \dfrac{1}{2}$

Where on the Unit Circle does $sin = \dfrac{1}{2}$

<em>Hint: sin is only positive in Quadrants I and II</em>

$\text{Answer: at}\ \dfrac{\pi}{6}\ (30^o)\ \text{and at}\ \dfrac{5\pi}{6}\ (150^o)\ \text{and all rotations of}\ 2\pi \ (360^o)$

$\text{In radians:}\ t = \dfrac{\pi}{6} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{6} + 2\pi n$

In degrees: t = 30° + 360n and 150° + 360n

******************************************************************************

$7)\quad tan (t) = -\sqrt3$

Where on the Unit Circle does $\dfrac{sin}{cos} = \dfrac{-\sqrt3}{1}\ or\ \dfrac{\sqrt3}{-1}\quad \rightarrow \quad (1,-\sqrt3)\ or\ (-1, \sqrt3)$

<em>Hint: sin and cos are only opposite signs in Quadrants II and IV</em>

$\text{Answer: at}\ \dfrac{2\pi}{3}\ (120^o)\ \text{and at}\ \dfrac{5\pi}{3}\ (300^o)\ \text{and all rotations of}\ 2\pi \ (360^o)$

$\text{In radians:}\ t = \dfrac{2\pi}{3} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{3} + 2\pi n$

In degrees: t = 120° + 360n and 300° + 360n

4 0

2 years ago

Keith ate 2/6 of a candy bar. if there were 6 pieces total,what fraction of the candy bar left?

Effectus [21]

<span>If there were 6 pieces total and Keith ate two of them.
So 6/6 - 2/6 = 4/6 or 2/3.</span>

3 0

2 years ago