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ziro4ka [17]
3 years ago
11

Which shows the correct solution of the equation, when ?

Mathematics
2 answers:
ch4aika [34]3 years ago
6 0
When the equation is ax+b which is y =ax+b
SSSSS [86.1K]3 years ago
5 0

Answer:

C a=60

Step-by-step explanation:

2/3(30)=20 subtract 20 from both sides   50-20=30     1/2a=30   multiply 2 from both sides.   a=60      

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Help pls, answer choices listed!
weeeeeb [17]

Answer:

I believe it is answer B if it's wrong I'm very sorry/

Step-by-step explanation:

6 0
3 years ago
. How can you determine if two ratios are equivalent. Be sure to include examples to support you response.
BabaBlast [244]

Answer:

By multiplying each ratio by the second number of the other ratio, you can determine if they are equivalent. Multiply both numbers in the first ratio by the second number of the second ratio. For example, if the ratios are 3:5 and 9:15, multiply 3 by 15 and 5 by 15 to get 45:75.

7 0
3 years ago
When 7/8x^2-3/4 x is subtracted from 5/8x^2-1/4x+2 the difference is?
artcher [175]
7/8x²-3/4x - 5/8x²-1/4x+2

x²/4-x+2
4 0
3 years ago
Read 2 more answers
HELLLLLLLLLPPPPPPP The difference of two numbers is 8 and their quotient is 2
Dmitriy789 [7]

Answer:

x-y =8

x/y=3

=> 3y-y=8

=> 2y = 8

=> y = 4

=> x = 4+8 =12

3 0
3 years ago
Find the equation of the parabola with focus (5, 1) and directrix y = -1.
Mumz [18]
Check the picture below.

since the focus point is above the directrix, that simply means the parabola is a vertical one, and therefore the square variable is the "x".

keeping in mind that, there's a distance "p" from the vertex to either the focus point or the directrix, that puts the vertex half-way between those fellows, in this case at 0, right between 1 and -1, as you see in the picture, and the parabola looks like so.

since the parabola is opening upwards, the value for "p" is positive, thus

\bf \textit{parabola vertex form with focus point distance}
\\\\
\begin{array}{llll}
4p(x- h)=(y- k)^2
\\\\
\boxed{4p(y- k)=(x- h)^2}
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ( h, k)\\\\
 p=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------\\\\
\begin{cases}
h=5\\
k=0\\
p=1
\end{cases}\implies 4(1)(y-0)=(x-5)^2
\\\\\\
4y=(x-5)^2\implies  y=\cfrac{1}{4}(x-5)^2

8 0
3 years ago
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