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qaws [65]
3 years ago
7

(03.03 MC)

Mathematics
1 answer:
notsponge [240]3 years ago
4 0

Answer:

The slope of f(x) is 10 and the slope of g(x) is 5; g(x) has the greater y-intercept.

To find the slope of f(x), we use the slope formula:  m=(y₂-y₁)/(x₂-x₁) = (-1--11)/(0--1) = (-1+11)/(0+1) = 10/1 = 10.

To find the slope of g(x), we just look at the form it is in.  It is written in slope-intercept form, y=mx+b, where m is the slope.  The number in g(x) that would correspond to m is 5.

The y-intercept of f(x) is found by looking at the points.  Any y-intercept will have an x-coordinate of 0; the only point like this in the table is (0, -1) so the y-intercept is -1.

For g(x), we again look at the form y=mx+b.  The number that corresponds with b is the y-intercept; in this case, it is 1.  1>-1, so g(x) has the larger y-intercept.

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What is the length of the unknown
AVprozaik [17]
The answer is A.

Explanation:

Since this is a right triangle you can use The formula: leg squared plus leg squared = hypotenuse squared

So lets square the hypotenuse and first leg
And you get 85 hypotenuse and 49 leg

So just subtract 49 from 85 to get the other leg squares

You get 36 so now just square root it

Square root of 36 is 6

So the best answer is 6

Hope this helps some.
8 0
3 years ago
If <img src="https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20x%20%3D%20log_%7Ba%7D%28bc%29" id="TexFormula1" title="\rm \: x = log_{a}(
timama [110]

Use the change-of-basis identity,

\log_x(y) = \dfrac{\ln(y)}{\ln(x)}

to write

xyz = \log_a(bc) \log_b(ac) \log_c(ab) = \dfrac{\ln(bc) \ln(ac) \ln(ab)}{\ln(a) \ln(b) \ln(c)}

Use the product-to-sum identity,

\log_x(yz) = \log_x(y) + \log_x(z)

to write

xyz = \dfrac{(\ln(b) + \ln(c)) (\ln(a) + \ln(c)) (\ln(a) + \ln(b))}{\ln(a) \ln(b) \ln(c)}

Redistribute the factors on the left side as

xyz = \dfrac{\ln(b) + \ln(c)}{\ln(b)} \times \dfrac{\ln(a) + \ln(c)}{\ln(c)} \times \dfrac{\ln(a) + \ln(b)}{\ln(a)}

and simplify to

xyz = \left(1 + \dfrac{\ln(c)}{\ln(b)}\right) \left(1 + \dfrac{\ln(a)}{\ln(c)}\right) \left(1 + \dfrac{\ln(b)}{\ln(a)}\right)

Now expand the right side:

xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} \\\\ ~~~~~~~~~~~~+ \dfrac{\ln(c)\ln(a)}{\ln(b)\ln(c)} + \dfrac{\ln(c)\ln(b)}{\ln(b)\ln(a)} + \dfrac{\ln(a)\ln(b)}{\ln(c)\ln(a)} \\\\ ~~~~~~~~~~~~ + \dfrac{\ln(c)\ln(a)\ln(b)}{\ln(b)\ln(c)\ln(a)}

Simplify and rewrite using the logarithm properties mentioned earlier.

xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} + \dfrac{\ln(a)}{\ln(b)} + \dfrac{\ln(c)}{\ln(a)} + \dfrac{\ln(b)}{\ln(c)} + 1

xyz = 2 + \dfrac{\ln(c)+\ln(a)}{\ln(b)} + \dfrac{\ln(a)+\ln(b)}{\ln(c)} + \dfrac{\ln(b)+\ln(c)}{\ln(a)}

xyz = 2 + \dfrac{\ln(ac)}{\ln(b)} + \dfrac{\ln(ab)}{\ln(c)} + \dfrac{\ln(bc)}{\ln(a)}

xyz = 2 + \log_b(ac) + \log_c(ab) + \log_a(bc)

\implies \boxed{xyz = x + y + z + 2}

(C)

6 0
2 years ago
I dont understand this question, and needs to be done by tomorrow morning please help​
gladu [14]

Answer:

ABCD and EFGH

ABCD and PQRS (or EFGH and PQRS)

Dilate by a scale factor of 3

Step-by-step explanation:

Congruent means they have the same shape and size.

Similar means they have the same shape, but not necessarily the same size.

The orientation (rotation angle) or position do not matter.

EFGH is reflected and rotated, so it maintains the same shape and size as ABCD.  Therefore, they are congruent.

PQRS is scaled and translated, so it has the same shape, but different size than ABCD.  Therefore, they are similar but not congruent.

Also, PQRS is similar to EFGH, but not congruent.

To make EFGH congruent to PQRS, we need to make it the same size.  So we need to scale EFGH by a factor of 3.

3 0
3 years ago
For a system of linear equations, the solution for the system is__
amm1812
For a system of linear equations, the solution for the system is____.

Answer: The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair (4,7) is the solution to the system of linear equations.
Picture:

8 0
2 years ago
Consider the differential equation y'' − y' − 30y = 0. Verify that the functions e−5x and e6x form a fundamental set of solution
Alex Ar [27]

Answer:

Step-by-step explanation:

We have to take the derivatives for both functions and replace in the differential equation. Hence

for y=e^{-5x}:

y(x)=e^{-5x}\\y'(x)=-5e^{-5x}\\y''(x)=25e^{-5x}\\

for y=e^{6x}:

y(x)=e^{6x}\\y'(x)=6e^{6x}\\y''(x)=36e^{6x}\\

Now we replace in the differential equation  y'' − y' − 30y = 0

for y=e^{-5x}:

25e^{-5x}+5e^{-5x}-30e^{-5x}=0\\25+5-30=0

for y=e^{6x}:

36e^{6x}-6e^{6x}-30=0\\36-6+30=0

Now, to know if both function are linearly independent we calculate the Wronskian

W(f,g)=fg'-f'g

W(e^{-5x},e^{6x})=(e^{-5x})(6e^{6x})-(-5e^{-5x})(e^{6x})\neq 0

I hope this is useful for you

Best regard

7 0
3 years ago
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