Can the statement be written as a biconditional statement and why?????

Mathematics

1 answer:

Gwar [14]3 years ago

5 0

No. Writing it as a biconditional would mean that "If a number is odd, then it is a prime number" is also true. The original statement is true, but the converse is false.

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Solve |3k-2|=2|k+12|

Stella [2.4K]

Answer:

$k = 26$ or $k = -\dfrac{22}{5}$

Step-by-step explanation:

|3k - 2| = 2|k + 12|

$\dfrac{|3k - 2|}{|k + 12|} = 2$

$|\dfrac{3k - 2}{k + 12}| = 2$

$3k - 2 = 2(k + 12)$ or $3k - 2 = -2(k + 12)$

$3k - 2 = 2k + 24$ or $3k - 2 = -2k - 24$

$k = 26$ or $5k = -22$

$k = 26$ or $k = -\dfrac{22}{5}$

7 0

3 years ago

Which of the following is the equation for the graph shown?a. x^2/144+y^2/95=1b. x^2/144-y^2/95=1c. x^2/95+y^2/144=1d. x^2/95-y^

Andrews [41]

e follSOLUTION

Given the question in the image, the following are the solution steps to answer the question

STEP 1: Write the general equation of an ellipse

$\frac{\mleft(x-h\mright)^2}{a^2}+\frac{(y-h)^2}{b^2^{}}=1$

STEP 2: Identify the parameters

the length of the major axis is 2a

the length of the minor axis is 2b

$\begin{gathered} 2a=24,a=\frac{24}{2}=12 \\ 2b=20,b=\frac{20}{2}=10 \end{gathered}$

STEP 3: Get the equation of the ellipse

$\begin{gathered} By\text{ substitution,} \\ \frac{(x-h)^2}{a^2}+\frac{(y-h)^2}{b^2}=1 \\ \frac{(x-0)^2}{12^2}+\frac{(y-0)^2}{10^2}=1=\frac{x^2}{144}+\frac{y^2}{100}=1 \end{gathered}$

STEP 4: Pick the nearest equation from the options,

Hence, the equation of the ellipse in the image is given as: