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Montano1993 [528]
3 years ago
13

Will also give brainliest to whoever is right

Mathematics
1 answer:
bezimeni [28]3 years ago
4 0
Yo just get Slader for big ideas. When you get on choose your book and the unit and question then done
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Which answer shows this equation in standard form? 7 – 3(x – y) = –5x + 2 A.–8x + 3y = –5 B.–2x – 3y = 5 C. 2x + 3y = –5 D. 2x –
Effectus [21]
<span>The question is which answer does show the standard form of the equation: 7-3(x-y)=-5x+2. To see that let's transform our equation into standard form. Standard form of a linear equation is: ax+by=c. So our equation is: 7-3x+3y=-5x+2. Now we put the unknowns on the left and the numbers on the right: -3x+5x+3y=2-7 and we get 2x+3y=-5. And we see that the correct answer is C. </span>
7 0
3 years ago
Please need help asap!!
NNADVOKAT [17]
The answer is false, they are parallel not perpendicular
5 0
3 years ago
-2x = - 14 value of x
pishuonlain [190]

x = 7

isolate x by dividing both sides by - 2

x = \frac{- 14}{- 2} = 7


7 0
3 years ago
Read 2 more answers
NO LINKS!! Please help me with this problem​
Vera_Pavlovna [14]

Information : The given hyperbola is a horizontal hylerbola with its centre (3 , -5) and one of its focus at (9 , -5) and vertex at (7 , -5) and as we can see that the focus and vertex have same y - coordinates, it must have its Transverse axis on line y = - 5.

Now,

it's vertex is given, I.e (7 , -5)

so, length of semi transverse axis will be equal to distance of vertex from centre, i.e

  • a = 7 - 3 = 4 units

Now, it's focus can be represented as ;

\qquad \sf  \dashrightarrow \: (3 + ae,  - 5 )

so,

  • ae + 3 = 9

and we know, a = 4

\qquad \sf  \dashrightarrow \: 4e + 3 = 9

\qquad \sf  \dashrightarrow \: 4e = 6

\qquad \sf  \dashrightarrow \: e =  \cfrac{3}{2}

Now, let's find the measure of semi - conjugate axis (b)

\qquad \sf  \dashrightarrow \:  {b}^{2}  =  {a}^{2} ( {e}^{2}  - 1)

\qquad \sf  \dashrightarrow \:  {b}^{2}  = 16( \frac{9}{4}  - 1)

\qquad \sf  \dashrightarrow \:  {b}^{2}  = 16( \frac{9 - 4}{4}  )

\qquad \sf  \dashrightarrow \:  {b}^{2}  = 16( \frac{5}{4}  )

\qquad \sf  \dashrightarrow \:  {b}^{2}  = 20

\qquad \sf  \dashrightarrow \: b =   \sqrt{20}

So, it's time to write the equation of hyperbola, as we already have the values of a and b ~

\qquad \sf  \dashrightarrow \:  \cfrac{ {(x - h)}^{2} }{ {a}^{2} }  -  \cfrac{( {y  - k)}^{2} }{ {b}^{2} }  = 1

[ plug in the values, and h = x - coordinate of centre, and k = y - coordinate of centre ]

\qquad \sf  \dashrightarrow \:  \cfrac{ ({x-3)}^{2} }{ {16}^{} }  -  \dfrac{ {(y+5)}^{2} }{ { {20} }^{} }  = 1

6 0
2 years ago
Read 2 more answers
Are 1/8 and 12.5% equivalent ​QUICK
Nataliya [291]

Answer:

Step-by-step explanation:

Yes they are equivalent

Hope it helps!

*Gotta go fast*

6 0
3 years ago
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