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andrew-mc [135]
4 years ago
15

Choose the answer based on the most efficient method as presented in the lesson. If the first step in the solution of the equati

on -8 - 7x = -5x - 10 is "add 10," then what should the next step be?

Mathematics
1 answer:
aleksley [76]4 years ago
7 0
-8x-7x=-5x-10
Add or move the -10 to the other side
It becomes: -8+10-7x=-5x
Which then becomes: 2-7x=-5x
Add or move the -7x to the other side
It becomes: 2=-5x+7x
Which then becomes: 2=2x
(I solved it in the picture)


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Help please I need help
serg [7]
He could use the mode of 89.
6 0
3 years ago
HELP WITH NUMBER 13 please
amid [387]

Answer: QE = 10

Step-by-step explanation: To solve this problem, it's important to understand that the diagonals of a parallelogram bisect each other.

This means that E is the midpoint of diagonal SQ.

So we can setup the equation x² + 9x = 4x + 6.

To solve this polynomial equation, set it equal to zero first.

So we have x² + 5x - 6 = 0 and we get (x + 6)(x - 1) = 0

when we factor the left side of the equation.

So this means that x = -6 or x = 1.

However, -6 will give us a negative length when we plug it in

to find QE so this will not work.

However, plugging 1 in will give us 10 as a length so QE = 10.

4 0
2 years ago
HELP PLEASE and show work
larisa86 [58]

Answer:

\boxed{\stackrel{\Large\frown}{PQS}\:= 222°}

Explanation:

Since \overline{PR} is a diameter, arc \stackrel{\Large\frown}{PQR}\:= 180°.

Given that \angle{SUR} = 42° is a subtended central angle by two diameters,

\stackrel{\large\frown}{RS} \:= 42°.

Using the rule of arcs, \stackrel{\Large\frown}{PQR} + \stackrel{\large\frown}{RS} \: = \: \stackrel{\Large\frown}{PQS} →

180° + \: 42° = \: \stackrel{\Large\frown}{PQS}

222° = \: \stackrel{\Large\frown}{PQS}

\stackrel{\Large\frown}{PQS} \: = \: 222°

3 0
3 years ago
Find the roots of h(t) = (139kt)^2 − 69t + 80
Sonbull [250]

Answer:

The positive value of k will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80, roots are those values of t so that h(t) = 0. That is:

19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0 (1)

Roots are determined analytically by the Quadratic Formula:

t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}

t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }

The smaller root is t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }, and the larger root is t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }.

h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80 has one real root when \frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0. Then, we solve the discriminant for k:

\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}

k \approx \pm 0.028

The positive value of k will result in exactly one real root is approximately 0.028.

7 0
3 years ago
What <br> is (81m^6)^1/2 simplified
ANEK [815]
(81m^6)^1/2 = (3^4*1/2) (m^6*1/2) = 3²m³
7 0
3 years ago
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