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Xelga [282]
3 years ago
15

Find values for p and q so that the equation has infinitely many solutions.

Mathematics
1 answer:
GaryK [48]3 years ago
3 0
Px-2= 4x+x+9

First what you should do is subtract the 2 on both sides

You’ll get px = 4x+x+7 then add the x to 4x

Px= 4x^2 +7

Dived the x on the left on both sides like this px/x = 4x^2+7/x

P= 4x+7
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3.5 pounds $1.19 per pound how much did she spend? to the nearest cent
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4.16 I think if I did the math right
6 0
3 years ago
Read 2 more answers
22÷1/5 (urgent) I need this ​
castortr0y [4]

Answer:

110

Step-by-step explanation:

easy matth to me

tip:22÷1/5 is 22x5

6 0
3 years ago
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Prove that the diagonals of a parallelogram bisect each other​
Nady [450]

Answer:

[ See the attached picture ]

The diagonals of a parallelogram bisect each other.

✧ Given : ABCD is a parallelogram. Diagonals AC and BD intersect at O.

✺ To prove : AC and BD bisect each other at O , i.e AO = OC and BO = OD.

Proof :\begin{array}{ |c| c |  c |  } \hline \tt{SN}& \tt{STATEMENTS} & \tt{REASONS}\\ \hline 1& \sf{In  \: \triangle ^{s}  \:AOB \: and \: COD  } \\  \sf{(i)}&  \sf{ \angle \: OAB =  \angle \: OCD\: (A)}& \sf{AB \parallel \: DC \: and \: alternate \: angles} \\  \sf{(ii)} &\sf{AB = DC(S)}& \sf{Opposite \: sides \: of \: a \: parallelogram} \\  \sf{(iii)} &\sf{ \angle \: OBA=  \angle \: ODC(A)} &\sf{AB \parallel \:DC \: and \: alternate \: angles} \\  \sf{(iv)}& \sf{ \triangle \:AOB\cong \triangle \: COD}& \sf{A.S.A \: axiom}\\ \hline 2.& \sf{AO = OC \: and \: BO = OD}& \sf{Corresponding \: sides \: of \: congruent \: triangle}\\ \hline 3.& \sf{AC \: and \: BD \: bisect \: each \: other \: at \: O}& \sf{From \: statement \: (2)}\\ \\ \hline\end{array}.          Proved ✔

♕ And we're done! Hurrayyy! ;)

# STUDY HARD! So, Tomorrow you can answer people like this , " Dude , I just bought this expensive mobile phone but it is not that expensive for me" [ - Unknown ] :P

☄ Hope I helped! ♡

☃ Let me know if you have any questions! ♪

\underbrace{ \overbrace  {\mathfrak{Carry \: On \: Learning}}} ☂

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

5 0
3 years ago
E
AlexFokin [52]

For this case we have that by definition, the point-slope equation of a line is given by:

y-y_ {0} = m (x-x_ {0})

Where:

m: It's the slope

(x_ {0}, y_ {0}): It is a point through which the line passes

According to the statement data we have:

m = 5\\(x_ {0}, y_ {0}): (- 8,1)

Substituting we have:

y-1 = 5 (x - (- 8))

By law of signs of multiplication we have to:

- * - = +\\y-1 = 5 (x + 8)

Answer:

The requested equation is: y-1 = 5 (x + 8)

5 0
3 years ago
2) Fill in an angle measure greater than 100 degrees for the missing angle in yellow below. You will fill
timama [110]

Answer:

m∠a = 110°

m∠b = 110°

m∠c = 70°

m∠d = 70°

m∠i = 110°

m∠h = 110°

m∠j = 70°

m∠g = 110°

m∠f = 70°

m∠e = 70°

m∠k = 57°

m∠m = 70°

m∠l = 167°

Step-by-step explanation:

Let the measure of the yellow angle = 110°

Therefore;

m∠a = 110°  by vertically opposite angles

m∠b = m∠a = 110° by opposite interior angles of a parallelogram

m∠c = 180° - 110° = 70° by same side interior angles

m∠d = m∠c = 70° by opposite interior angles of a parallelogram

m∠i and m∠c are supplementary angles, therefore, m∠i = 110°

m∠h = m∠i = 110° by vertically opposite angles

m∠j = m∠c = 70° by vertically opposite angles

m∠g and m∠d are supplementary angles, therefore, m∠g = 110°

m∠f = m∠d = 70° by alternate interior angles

m∠f = m∠e = 70° by vertically opposite angles

m∠k = 180° - (m∠h + 13°) = 180° - (110° + 13°) = 57°

m∠b and m∠m are supplementary angles, therefore, m∠m = 70°

m∠l and 13° are supplementary angles, therefore, m∠l = 167°

4 0
3 years ago
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