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DaniilM [7]
3 years ago
10

100 POINTS NEEDED RIGHT NOW! On a graph plot the points where the following function crosses the x-axis and y-axis g(x) = -5^x+5

Mathematics
1 answer:
Aloiza [94]3 years ago
5 0

Answer:

see below

Step-by-step explanation:

X-AXIS

when a graph intersects the x-axis, the y value is always 0.  

Ex.  Points (1,0), (-5,0), (-3,0), (6,0), etc. are all on the x-axis because y = 0.  

to find where the graph crosses the x-axis, let y or g(x) = 0

0 = -5^x + 5  subtract 5 from both sides

-5 = -5^x       if x is 1, that would produce -5 = -5

(1,0) x axis

Y-AXIS

when a graph intersects the y-axis, the x value is always 0.

Ex.  Points (0, 1), (0, -2), (0, 3), etc. all lie on the y-axis because x = 0.

to find where teh graph crosses the y-axis, let x = 0

does the exponent "x" apply to (-5) or just 5?  

-5^{x}  or  (-5)^{x}

(-5)^0 + 5    any number with a 0 power is 1

1 + 5 = 6

(0, 6)

or -5^0 + 5

       -1  +  5 = 4

(0, 4)

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Write the domain and range of the graph below…
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\qquad\qquad\huge\underline{{\sf Answer}}

Let's see what's the domain and range for the given function ~

\qquad \tt \dashrightarrow \:domain \in [ -5 , 4 ]

or

\qquad \tt \dashrightarrow \: - 5 \leqslant x \leqslant 4

Because domain = all values of x between its maximum (4) and minimum (-5) value [ including them ]

\qquad \tt \dashrightarrow \:range \in[ -4 , 9 ]

or

\qquad \tt \dashrightarrow \: - 4 \leqslant y \leqslant 9

Because range = all values of y between its maximum (9) and minimum (-4) value [ including them ]

4 0
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The diagram shows a right-angled triangle.
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Answer:

h = 9.87 cm

Step-by-step explanation:

Given that,

Base, b = 8 cm

Height = h

Angle, \theta=51^{\circ}

We need to find the value of h. We can use trigonometry to find h.

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7 0
4 years ago
When 2(3/5x+2 3/4y-1/4x-1 1/2y+3) is simplified, what is the resulting expression?
erik [133]

Answer:

Option B is correct.

Step-by-step explanation:

2(\frac{3}{5}x+ 2\frac{3}{4}y-\frac{1}{4}x-1\frac{1}{2}y+3)

We need to solve the above expression.

First Convert Mix fraction into improper fraction

2(\frac{3}{5}x+ 2\frac{3}{4}y-\frac{1}{4}x-1\frac{1}{2}y+3)\\=2(\frac{3}{5}x+ \frac{11}{4}y-\frac{1}{4}x-\frac{3}{2}y+3)

Combining the like terms

=2(\frac{3}{5}x-\frac{1}{4}x+ \frac{11}{4}y-\frac{3}{2}y+3)

Solving like terms

=2(\frac{3x*4-5x}{20}+ \frac{11y-3y*2}{4}+3)\\=2(\frac{12x-5x}{20}+ \frac{11y-6y}{4}+3)\\=2(\frac{7x}{20}+ \frac{5y}{4}+3)

Multiply each term with 2

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=\frac{7x}{10}+ 2\frac{1}{2}y+6

So, Option B is correct.

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