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brilliants [131]
3 years ago
7

Which equation represents a line which is parallel to the line y=8/5x−1?

Mathematics
1 answer:
Murljashka [212]3 years ago
8 0

Answer:

y=8/5x+2

Step-by-step explanation:

if parallel, the slopes should be the same

slope = 8/5

so the equation could be y=8/5x+2, or anything that is moved across the y axis with that slope

You might be interested in
In right △ABC , the right angle is at C, m∠A=30∘ , and AC=5√ 5 units.
Makovka662 [10]

Answer:

5\sqrt{15}+5\sqrt{5}

Step-by-step explanation:

The core uses the value of the Pythagorean theorem and special angles.

6 0
3 years ago
Simplify the expression: -7d+5(10d+-6)
Alexandra [31]

Answer:

-7d+50d+-30= 43d+-30

  1. Use distributive property when there is parentheses and a number infront.
  2. combine like terms. ONLY if they have the same variable.
  3. then just put them in an expression.
3 0
3 years ago
How do I solve #7? do i add all the angles together like a normal triangle and make it equal to 180?
kobusy [5.1K]

Answer:

126 degrees

Step-by-step explanation:

Triangle: y = 180 - (5x + 4x)

Straight angle: y = 180 - (4x + 30)

180 - (5x + 4x) = 180 - (4x + 30)

5x + 4x = 4x + 30

5x = 30

x = 6

Angle:

y = 180 - (5x + 4x)

y = 180 - 9x

y = 180 - 9(6)

y =  126

7 0
2 years ago
The function f(x) = x^2 - 2x + 8 is transformed such that g(x) = f(x-2). Find the vertex of g(x).
Serjik [45]

Answer:

Step-by-step explanation:

Since g(x) is f(x-2), every time there is a "x" in the f(x) equation, substitute "x-2". So g(x) would be (x-2)^2-2(x-2)+8. Use the method of trial and error: substitute x as 1,-1, and 3, until you found the right vertex. The answer is B (3,7).

7 0
3 years ago
Rationalize the denominator of sqrt -49 over (7 - 2i) - (4 + 9i)
zubka84 [21]
\sqrt{ \frac{-49}{(7-2i)-(4+9i) } } 


This one is quite the deal, but we can begin by distributing the negative on the denominator and getting rid of the parenthesis:

\frac{ \sqrt{-49}}{7-2i-4-9i}

See how the denominator now is more a simplification of like terms, with this I mean that you operate the numbers with an "i" together and the ones that do not have an "i" together as well. Namely, the 7 and the -4, the -2i with the -9i.
Therefore having the result: 

\frac{ \sqrt{-49} }{3-11i}

Now, the \sqrt{-49} must be respresented as an imaginary number, and using the multiplication of radicals, we can simplify it to \sqrt{49}  \sqrt{-1}
This means that we get the result 7i for the numerator.

\frac{7i}{3-11i}

In order to rationalize this fraction even further, we have to remember an identity from the previous algebra classes, namely: x^2 - y^2 =(x+y)(x-y)
The difference of squares allows us to remove the imaginary part of this fraction, leaving us with a real number, hopefully, on the denominator.

\frac{7i (3+11i)}{(3-11i)(3+11i)}

See, all I did there was multiply both numerator and denominator with (3+11i) so I could complete the difference of squares.
See how (3-11i)(3+11i)= 3^2 -(11i)^2 therefore, we can finally write:

\frac{7i(3+11i)}{3^2 - (11i)^2 }

I'll let you take it from here, all you have to do is simplify it further.
The simplification is quite straightforward, the numerator distributed the 7i. Namely the product 7i(3+11i) = 21i+77i^2.
You should know from your classes that i^2 = -1, thefore the numerator simplifies to -77+21i
You can do it as a curious thing, but simplifying yields the result:
\frac{-77+21i}{130}
7 0
3 years ago
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