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yan [13]
3 years ago
11

How to solve 6x+3=5x+10

Mathematics
2 answers:
laila [671]3 years ago
7 0
6x+3=5x+10 \\\\ 6x-5x=10-3 \\\\ \boxed{x=7}
lakkis [162]3 years ago
4 0
If you follow these simple steps, it will make sense.
6x-5x+3=5x-5x+10
x+3=10
x+3-3=10-3
x=7
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Solve the system of equations by substitution. What is the solution for x?
insens350 [35]

Answer:

x=0.875

Step-by-step explanation:

2×+y=1

y=0.5

4×+2y=-1

4×+2.5=-1

1+2.5=4x

3.5=4x

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For the direct variation such that when y = 2 then x = 3 , find the constant of variation ( k) and then find the value of y when
Sergeu [11.5K]

Answer:

k is 2/3 and y is -1/3 when x is -0.5.

Step-by-step explanation:

The direct variation relationship is y = kx, where k is the const. of var.

Subbing 3 for x and 2 for y, 2 = 3k, or k = 2/3.

Now, if x = -0.5, y = (2/3)(-1/2) = -1/3

k is 2/3 and y is -1/3 when x is -0.5.

8 0
2 years ago
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chubhunter [2.5K]

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4 0
3 years ago
A line tangent to the curve f(x)=1/(2^2x) at the point (a, f(a)) has a slope of -1. What is the x-intercept of this tangent?
kirza4 [7]

Answer:

x-intercept = 0.956

Step-by-step explanation:

You have the function f(x) given by:

f(x)=\frac{1}{2^{2x}}   (1)

Furthermore you have that at the point (a,f(a)) the tangent line to that point has a slope of -1.

You first derivative the function f(x):

\frac{df}{dx}=\frac{d}{dx}[\frac{1}{2^{2x}}]  (2)

To solve this derivative you use the following derivative formula:

\frac{d}{dx}b^u=b^ulnb\frac{du}{dx}

For the derivative in (2) you have that b=2 and u=2x. You use the last expression in (2) and you obtain:

\frac{d}{dx}[2^{-2x}]=2^{-2x}(ln2)(-2)

You equal the last result to the value of the slope of the tangent line, because the derivative of a function is also its slope.

-2(ln2)2^{-2x}=-1

Next, from the last equation you can calculate the value of "a", by doing x=a. Furhtermore, by applying properties of logarithms you obtain:

-2(ln2)2^{-2a}=-1 \\\\2^{2a}=2(ln2)=1.386\\\\log_22^{2a}=log_2(1.386)\\\\2a=\frac{log(1.386)}{log(2)}\\\\a=0.235

With this value you calculate f(a):

f(a)=\frac{1}{2^{2(0.235)}}=0.721

Next, you use the general equation of line:

y-y_o=m(x-x_o)

for xo = a = 0.235 and yo = f(a) = 0.721:

y-0.721=(-1)(x-0.235)\\\\y=-x+0.956

The last is the equation of the tangent line at the point (a,f(a)).

Finally, to find the x-intercept you equal the function y to zero and calculate x:

0=-x+0.956\\\\x=0.956

hence, the x-intercept of the tangent line is 0.956

5 0
2 years ago
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