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3 years ago

HHHHHHHEEEEEEEEEEEEEEEEEEELLLLLLLLLLLLLLLLLLLLLPPPPPPPP

Mathematics

1 answer:

Furkat [3]3 years ago

8 0

Answer:

120

Step-by-step explanation:

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Carlos went to gas station A and paid $54 for 25 gallons. Sasha went to gas

RideAnS [48]

Answer: Sasha

Step-by-step explanation:

carlos- 54 divided by 25 equals 2.25 dollars for each gallon

sasha- 31.25 divided by 15 equals 2.083333(3 repeats)

They are both really close on price, but sasha's is cheaper

5 0

3 years ago

How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y

son4ous [18]

Solution:

Given:

$x^2=6y$

Part A:

The vertex of an up-down facing parabola of the form;

$\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}$

Rewriting the equation given;

$\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\ \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\ \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\ \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\ \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}$

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

$\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}$

Therefore, the focus is;

$(0,\frac{3}{2})$

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

$\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}$

Therefore, the directrix is;

$y=-\frac{3}{2}$

3 0

1 year ago

Select the postulate of equality or inequality that is illustrated.

Airida [17]

Hello,
"<" is transitive.
a<b
b<2 ==>a<b<2==>a<2

4 0

3 years ago

Read 2 more answers

BRAINLIEST!!!!!!!

Murrr4er [49]

All the angles inside a quadrilateral add to 360 degrees.
But first, we should solve for x
Opposite angles add up to 180 degrees.
So we'll use angles B and D
x + 10 + x + 24 = 180
combine like terms
2x + 34 = 180
subtract 34 from both sides
2x = 146
divide both sides by 2
x = 73

Now let's solve for the angles
Angle D: x + 24 = 97
Angle B: x + 10 = 83
Angle A: x + 15 = 88

Opposite angles add up to 180
we know that angle A is 88
C + 88 = 180
subtract 88 from both sides
C = 92

Hope this helps!

4 0

3 years ago

Let f(x)equals=33xminus−1, h(x)equals=startfraction 7 over x plus 5 endfraction 7 x+5 . find (hcircle◦f)(66).

Ronch [10]

$\bf \begin{cases}
f(x)=3x-1\\
h(x)=\cfrac{7}{x+5}\\\\
(h\circ f)(x)=h(~~f(x)~~)
\end{cases}
\\\\\\
f(66)=3(66)-1\implies \boxed{f(66)=197}
\\\\\\
\stackrel{h(~~f(x)~~)}{h(~~f(66)~~)}\implies \stackrel{h(~~f(66)~~)}{h\left(~~\boxed{197}~~ \right)}=\cfrac{7}{\boxed{197}+5}\implies h\left(~~\boxed{197}~~ \right)=\cfrac{7}{202}$

7 0

3 years ago