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3 years ago

12

Solve these equations. Show solutions on a number line. |x+26| = -5 (2 answers)

1 answer:

Evgen [1.6K]3 years ago

4 0

Answer:

Step-by-step explanation:

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Has to be turned in soon at 9:00 it's 7:50

disa [49]

Answer:

A

Step-by-step explanation:

b must always be greater than 0 since 12b-15 must always be greater than 21. You can simplify the ineqaulity to see things easier; 12b>36. Again, this ineqaulity can be simplified; b>3. A is the closest answer here, but there should be an open interval at 3 for A to be correct.

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2 years ago

If EF=3x-12, FG=4x-15, and EG=29, enter the values of x, EF, and FG.

anzhelika [568]

Answer:

X =8

EF=12

FG=17

Step-by-step explanation:

4 0

2 years ago

HELP PLEASE HELP ITS THE LAST QUESTION

andrew-mc [135]

C and d hope it help

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2 years ago

Please help confused?

coldgirl [10]

All you've got to do is write an equation. EX 2n-1

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2 years ago

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NO LINKS!! Please help me with this problem

iogann1982 [59]

${\qquad\qquad\huge\underline{{\sf Answer}}}$

The given figure shows a vertical hyperbola with its centre at origin, and as we observe the figure, we can conclude that :

Length of transverse axis is :

$\qquad \sf \dashrightarrow \: 2b = 12$

$\qquad \sf \dashrightarrow \: b = 6$

length of conjugate axis is :

$\qquad \sf \dashrightarrow \: 2a = 8$

$\qquad \sf \dashrightarrow \: a = 4$

Equation of hyperbola ~

$\qquad \sf \dashrightarrow \: \cfrac{ {y}^{2} }{ {b}^{2} } - \cfrac{ {x}^{2} }{ {a}^{2} } = 1$

plug in the values ~

$\qquad \sf \dashrightarrow \: \cfrac{ {y}^{2} }{ {6}^{2} } - \cfrac{ {x}^{2} }{ {4}^{2} } = 1$

$\qquad \sf \dashrightarrow \: \cfrac{ {y}^{2} }{ {36}^{} } - \cfrac{ {x}^{2} }{ {16}^{} } = 1$

5 0

1 year ago

Read 2 more answers