All categories

3 years ago

Which statement is NOT true about this figure?

Mathematics

1 answer:

Lunna [17]3 years ago

8 0

Answer:

od I'd the correct answer

You might be interested in

A water pitcher holds 3.5 quarts of water. How many liters of water does it hold?

DiKsa [7]

3.5 quarts = 3.31 liters

6 0

3 years ago

Read 2 more answers

−7x 2 +2x−1=0minus, 7, x, squared, plus, 2, x, minus, 1, equals, 0. I need this quick

BlackZzzverrR [31]

Answer:

-7x² + 2x - 1 = 0

x = - (1 - 2√2)/7

x = - (1 + 2√2)/7

3 0

4 years ago

A number line goes from negative 5 to positive 5. Point D is at negative 4 and point E is at positive 5. A line is drawn from po

Mrac [35]

Answer:

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

9x + 8 y = -6 -9x - 9y - 9 Elimination

kolezko [41]

Answer:

x = 2, y = -3

Step-by-step explanation:

Add them together and get

9x + 8y - 9x - 9y = -6 + 9

-y = 3

so y = -3.

Sub it into the first equation and get

9x + 8(-3) = -6

9x - 24 = -6

9x = -6 + 24 = 18

x = 18/9 = 2

6 0

2 years ago

A sine function had an amplitude of 3, period of 6pi, horizontal shift of 3pi/2, & vertical shift of -1.

Simora [160]

Answer: $\bold{y=\dfrac{1}{2}}$

Step-by-step explanation:

f(x) = A sin (Bx - C) + D

amplitude = |A|
period = $\dfrac{2\pi}{B}$
phase shift = $\dfrac{C}{B}$
vertical shift = D

A

amplitude of 3 is given so 3 = |A| → A = ± 3, since it is stated that this is a positive function, then A = 3

B

period of 6π is given so $6\pi=\dfrac{2\pi}{B}\quad \rightarrow \quad B=\dfrac{2\pi}{6\pi}\quad \rightarrow \quad B=\dfrac{1}{3}$

C

$\text{phase shift is given as}\ \dfrac{3\pi}{2}\ \text{so}\ \dfrac{3\pi}{2}=\dfrac{C}{\frac{1}{3}}\quad \rightarrow\quad \dfrac{(\frac{1}{3})3\pi}{2}=C\quad \rightarrow\quad \dfrac{\pi}{2}=C$

D

vertical shift of -1 is given so -1 = D

Now, substitute the values of A, B, C, and D into the formula (above):

$f(x) = 3\ sin \bigg(\dfrac{1}{3}x - \dfrac{\pi}{2}\bigg) - 1$

Next, solve when x = 2π

$f(2\pi) = 3\ sin \bigg(\dfrac{1}{3}(2\pi) - \dfrac{\pi}{2}\bigg) - 1$

$= 3\ sin \bigg(\dfrac{2\pi}{3} - \dfrac{\pi}{2}\bigg) - 1$

$= 3\ sin \bigg(\dfrac{4\pi}{6} - \dfrac{3\pi}{6}\bigg) - 1$

$= 3\ sin \bigg(\dfrac{\pi}{6}\bigg) - 1$

$= 3\ \bigg(\dfrac{1}{2}\bigg) - 1$

$=\dfrac{3}{2}-\dfrac{2}{2}$

$=\dfrac{1}{2}$

6 0

3 years ago