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Sonja [21]
2 years ago
7

PLEASE HELP ME QUICK!!

Mathematics
1 answer:
9966 [12]2 years ago
3 0

Answer: yes kevin skied as long as lori

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F(x) = <img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%2B7%7D%20-%5Csqrt%7Bx%5E2%2B2x-15%7D" id="TexFormula1" title="\sqrt{x+7} -\
elena-s [515]

Answer:

x >= -7  ................(1a)

x >= 3   ...............(2a1)

Step-by-step explanation:

f(x) =  \sqrt{x+7}-\sqrt{x^2+2x-15}  .............(0)

find the domain.

To find the (real) domain, we need to ensure that each term remains a real number.

which means the following conditions must be met

x+7 >= 0  .....................(1)

AND

x^2+2x-15 >= 0 ..........(2)

To satisfy (1),  x >= -7  .....................(1a) by transposition of (1)

To satisfy (2), we need first to find the roots of (2)

factor (2)

(x+5)(x-3) >= 0

This implis

(x+5) >= 0 AND (x-3) >= 0.....................(2a)

OR

(x+5) <= 0 AND (x-3) <= 0 ...................(2b)

(2a) is satisfied with x >= 3   ...............(2a1)

(2b) is satisfied with x <= -5 ................(2b1)

Combine the conditions (1a), (2a1), and (2b1),

x >= -7  ................(1a)

AND

(

x >= 3   ...............(2a1)

OR

x <= -5 ................(2b1)

)

which expands to

(1a) and (2a1)   OR  (1a) and (2b1)

( x >= -7 and x >= 3 )  OR ( x >= -7 and x <= -5 )

Simplifying, we have

x >= 3  OR ( -7 <= x <= -5 )

Referring to attached figure, the domain is indicated in dark (purple), the red-brown and white regions satisfiy only one of the two conditions.

3 0
3 years ago
Read 2 more answers
#78 please explain work
kenny6666 [7]

Answer:

$0.40c + $0.20h = $11

Step-by-step explanation:

c = chocolate pieces

h = hard candy

$0.40c + $0.20h = $11

5 0
3 years ago
Find the equation of a parabola with a vertical axis and its vertex at the origin and passing through the point (-2, 3)
vredina [299]

a vertical axis, I assume it means a vertical axis of symmetry, thus it'd be a vertical parabola, like the one in the picture below.

\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} y=a(x- h)^2+ k\qquad \qquad \leftarrow vertical\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=0\\ k=0 \end{cases}\implies y=a(x-0)^2+0 \\\\\\ \textit{we also know that } \begin{cases} x=-2\\ y=3 \end{cases}\implies 3=a(-2-0)^2+0\implies 3=4a \\\\\\ \cfrac{3}{4}=a~\hspace{10em}y=\cfrac{3}{4}(x-0)^2+0\implies \boxed{y=\cfrac{3}{4}x^2}

8 0
3 years ago
At one time, the Texas Junior College Teachers Association annual conference was held in Austin. At that time a taxi ride in Aus
slavikrds [6]
$6.75 would be your answer.
5 0
3 years ago
I need to know the answer
solniwko [45]
X-3/8=-1/4
+3/8
x=-1/4+3/8
make both bottom numbers the same so you are able to add them
-1/4×2/2= -2/8
x=-2/8+3/8
x=1/8
4 0
3 years ago
Read 2 more answers
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