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o-na [289]
4 years ago
14

Please answer this question thank you for your help

Mathematics
1 answer:
vovangra [49]4 years ago
4 0
Hey!
What I did was I graphed them or you could set them equal to each other and use logs!
I graphed them and found the intersections. There were 2, but one was negative so I checked for a positive intersection and found one. 
The first blank is 7 and the other is 140!
Hope this helps!
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What is the solution to the system of equations? Use the substitution method. {y=2x+43x−6y=3 Enter your answer in the boxes
UNO [17]

Answer:

the solution is (-3, -2)

Step-by-step explanation:

Please, separate your equations with a (;) or a (.).  Better yet, write only one equation per line:

y=2x+4

3x−6y=3

Substitute 2x + 4 for y in the second equation:

3x - 6(2x + 4) = 3, or

3x - 12x - 24 = 3

Combining like terms, we get -9x = 27, and so x = -27/9, or x = -3.

Since we know that y=2x+4, we replace x in this equation with -3 and calculate y:

y = 2(-3) + 4 = -2

Then the solution is (-3, -2).

6 0
3 years ago
WHAT IS THE EQUATION OF THE LINES IN SLOPE -INTERCEPT FORM ?
Natasha2012 [34]

Answer:

y=x-5

Step-by-step explanation:

the y intercept is -5 and slope is 1

5 0
3 years ago
Read 2 more answers
Integration of dx/3sin^2 -4​
Scilla [17]

Answer:

= 1/2 tan^(-1)(2 cot(x)) + constant

Step-by-step explanation:

Take the integral:

integral1/(3 sin^2(x) - 4) dx

Multiply numerator and denominator of 1/(3 sin^2(x) - 4) by -sec^2(x):

= integral-(sec^2(x))/(4 sec^2(x) - 3 tan^2(x)) dx

Prepare to substitute u = tan(x). Rewrite -(sec^2(x))/(4 sec^2(x) - 3 tan^2(x)) using sec^2(x) = tan^2(x) + 1:

= integral-(sec^2(x))/(tan^2(x) + 4) dx

For the integrand -(sec^2(x))/(tan^2(x) + 4), substitute u = tan(x) and du = sec^2(x) dx:

= integral-1/(u^2 + 4) du

Factor out constants:

= - integral1/(u^2 + 4) du

Factor 4 from the denominator:

= - integral1/(4 (u^2/4 + 1)) du

Factor out constants:

= -1/4 integral1/(u^2/4 + 1) du

For the integrand 1/(u^2/4 + 1), substitute s = u/2 and ds = 1/2 du:

= -1/2 integral1/(s^2 + 1) ds

The integral of 1/(s^2 + 1) is tan^(-1)(s):

= -1/2 tan^(-1)(s) + constant

Substitute back for s = u/2:

= -1/2 tan^(-1)(u/2) + constant

Substitute back for u = tan(x):

= -1/2 tan^(-1)(tan(x)/2) + constant

Which is equivalent for restricted x values to:

Answer: = 1/2 tan^(-1)(2 cot(x)) + constant

6 0
3 years ago
How are equations and inequalities the same and how are they different
xenn [34]

Answer:

i know right

Step-by-step explanation:

boting jeese

6 0
4 years ago
Read 2 more answers
A line passes through the points (8, –1) and (–4, 2).
Yuki888 [10]

Step-by-step explanation:

With the given points, we can first find the gradient of the line:

gradient = (2+1)/(-4-8)

= -1/4

We also know that the formula of a straight line is

Y = mX+C, where m is gradient and C is y intercept.

Hence by substituting the coordinates (-4, 2) and the gradient -1/4, we get:

2 = (-1/4)(-4) + C

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(Actually, you can just draw a straight line connecting the 2 coordinates and find out the y intercept)

8 0
3 years ago
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