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3 years ago

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Mathematics

1 answer:

Dahasolnce [82]3 years ago

4 0

Where the rectangles

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Graph the Rarabola.---Plot five points on the parabola: the vertex, two points to the left of the vertex, and tvbutton2Explanati

valentinak56 [21]

Which parabola?

$y=-x^2$

Equation of the parabola

y - y1 = 4p(x - x1)

The Vertex of the parabola given is (0, 0) because it does not have the values of x1 and y1.

Then Vertex = (0,0)

-Look for two values of y to the left and two points to the right.

You can choose the points that you which

x y

-3 y = -(-3)^2 = -9

-1 y = -(-1)^2 = -1

0 y = -(0)^2 = 0

1 y = -(1)^2 = -1

3 y = -(3)^2 = -9

3 0

1 year ago

A taxi service charges a 10$ initial fee and an additional $1.25 per minute.

Len [333]

Y= 10 + 1.25x
This is the answer because 10 is the initial fee (you start with it) and 1.25 is added for each minute, with each minute being the variable

7 0

3 years ago

140 fluid ounces increased by 45%

Radda [10]

$100\%\ increased\ by\ 45\%=100\%+45\%=145\%\\\\method\ \#1\\\\p\%=\frac{p}{100}\to145\%=\frac{145}{100}=\frac{145:5}{100:5}=\frac{29}{20}\\\\145\%\ of\ 140=\frac{29}{20}\cdot140=29\cdot7=\boxed{203\ (ounces)}\leftarrow answer$

$method\ \#2\\\\\begin{array}{ccc}140&-&100\%\\x&-&145\%\end{array}\ \ \ cross\ multiply\\\\\\100\cdot x=140\cdot145\ \ \ \ |divide\ both\ sides\ by\ 100\\\\x=\frac{140\cdot145}{100}\\\\\boxed{x=203\ (ounces)}$

4 0

3 years ago

The sample space of a random experiment is {a, b, c, d, e} with probabilities 0.1, 0.1, 0.2, 0.4, and 0.2 respectively. Let A de

antoniya [11.8K]

Answer:

a) $P(A) =P(a)+P(b) +P(c)= 0.1+0.1+0.2 = 0.4$

b) $P(B) =P(c) +P(d)+P(e)=0.2+0.4+0.2=0.8$

c) $P(A') = 1-P(A) =1-0.4=0.6$

d) $P(A \cup B) =0.4 +0.8-0.2 =1.0$

e) The intersection between the set A and B is the element c so then we have this:

$P(A \cap B) = P(c) =0.2$

Step-by-step explanation:

We have the following space provided:

$S= [a,b,c,d,e]$

With the following probabilities:

$P(a) =0.1, P(b)=0.1, P(c) =0.2, P(d)=0.4, P(e)=0.2$

And we define the following events:

A= [a,b,c], B=[c,d,e]

For this case we can find the individual probabilities for A and B like this:

$P(A) = 0.1+0.1+0.2 = 0.4$

$P(B) =0.2+0.4+0.2=0.8$

Determine:

a. P(A)

$P(A) =P(a)+P(b) +P(c)= 0.1+0.1+0.2 = 0.4$

b. P(B)

$P(B) =P(c) +P(d)+P(e)=0.2+0.4+0.2=0.8$

c. P(A’)

From definition of complement we have this:

$P(A') = 1-P(A) =1-0.4=0.6$

d. P(AUB)

Using the total law of probability we got:

$P(A \cup B) =P(A) +P(B)-P(A \cap B)$

For this case $P(A \cap B) = P(c) =0.2$ , so if we replace we got:

$P(A \cup B) =0.4 +0.8-0.2 =1.0$

e. P(AnB)

The intersection between the set A and B is the element c so then we have this:

$P(A \cap B) = P(c) =0.2$

8 0

3 years ago

Which of the following would be an acceptable first step in simplifying the expression sinx/1-sinx

Paha777 [63]

$\bf \textit{difference of squares}
\\\\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)
\\\\\\
\textit{also recall that }sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\
-------------------------------$

$\bf \cfrac{sin(x)}{1-sin(x)}\implies \cfrac{sin(x)}{1-sin(x)}\cdot \cfrac{1+sin(x)}{1+sin(x)}\implies \stackrel{first~step}{\cfrac{sin(x)[1+sin(x)]}{[1-sin(x)][1+sin(x)]}}
\\\\\\
\cfrac{sin(x)[1+sin(x)]}{1^2-sin^2(x)}\implies \cfrac{sin(x)[1+sin(x)]}{cos^2(x)}
\\\\\\
\cfrac{sin(x)+sin^2(x)}{cos^2(x)}\implies \cfrac{sin(x)}{cos^2(x)}+ \cfrac{sin^2(x)}{cos^2(x)}
\\\\\\
\cfrac{sin(x)}{cos(x)}\cdot \cfrac{1}{cos(x)}+\cfrac{sin^2(x)}{cos^2(x)}\implies tan(x)sec(x)+tan^2(x)
\\\\\\
tan(x)[sec(x)+tan(x)]$

8 0

3 years ago

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