What are the coordinates of the x-intercept of the equation -5y=4-2x?

Mathematics

1 answer:

umka2103 [35]4 years ago

6 0

The x intercept of -5y=4-2x equals (2,0)

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What is the value of AAA when we rewrite 2^{x-6}+2^{x}2 x−6 +2 x 2, start superscript, x, minus, 6, end superscript, plus, 2, st

Nuetrik [128]

Answer:

$A=\frac{65}{64}\ or\ A=1.015625$

Step-by-step explanation:

Given:

The expression is given as:

$2^{x-6}+2^{x}$

The equivalent expression to the above expression is given as:

$A\cdot 2^{x}$

Now, simplifying the original expression using the law of indices:

$a^{m-n}=\frac{a^m}{a^n}$

So, $2^{x-6}=\frac{2^x}{2^6}$ . The expression becomes:

$=\frac{2^x}{2^6}+2^x$

Now, $2^x$ is a common factor to both the terms, so we factor it out. This gives,

$=2^x(\dfrac{1}{2^6}+1)\\\\=2^x(\frac{1}{64}+1)\\\\=2^x(\frac{1}{64}+\frac{64}{64})\\\\=2^x(\frac{1+64}{64})\\\\=2^x(\frac{65}{64})\\\\=(\frac{65}{64})\cdot 2^x$

Now, on comparing the simplified form with the equivalent expression, we conclude:

$A\cdot2^x=(\frac{65}{64})\cdot 2^x\\\\A=\frac{65}{64}\ or\ 1.015625$

Therefore, the value of 'A' is $A=\frac{65}{64}\ or\ A=1.015625$

7 0

3 years ago

Read 2 more answers

Please Help Me Im timed

Mrac [35]

A is the answer. The graph shows a decrease i the value as age increases.
Hope this helps!!! :-)

3 0

4 years ago

(Please show all work) 1. Find f(6) if f(x)= 1/2x + x - 4

dsp73

Answer:

$\large\boxed{1.\ f(6)=5\ or\ f(6)=20}\\\boxed{2.\ d=5}\\\boxed{3.\ vertex=(-3,\ -1)}$

Step-by-step explanation:

$1.\\\text{Put x = 6 to the equation of a function}\ f(x):\\\\\text{If}\ f(x)=\dfrac{1}{2}x+x-4\to f(6)=\dfrac{1}{2}(6)+6-4=3+6-4=5\\\\\text{If}\ f(x)=\dfrac{1}{2}x^2+x-4\to f(6)=\dfrac{1}{2}(6^2)+6-4=\dfrac{1}{2}(36)+2=18+2=20\\\\2.\\\text{The formula of a distance between two points:}\\\\d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\\text{We have the points (2, -4) and (5, -8). Substitute:}\\\\d=\sqrt{(5-2)^2+(-8-(-4))^2}=\sqrt{3^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5$

$3.\\\text{The vertex formula of a parabola:}\\\\y=a(x-h)^2+k\\\\(h,\ k)-vertex\\\\\text{We have}\ y=(x+3)^2-1=(x-(-3))^2-1\\\\\text{Therefore the vertex is:}\ (-3,\ -1).$