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CaHeK987 [17]
3 years ago
10

Solve 7x-6 > 8 and 3x + 4> 22 for x. Could you please help solve this.

Mathematics
2 answers:
vova2212 [387]3 years ago
5 0
7x-6>8
7x>14
x>2

3x+4>22
3x>18
x>6
Fiesta28 [93]3 years ago
4 0
7x-6 > 8 \\ \\7x>8+6 \\ \\7x>14 \ \ :7 \\ \\x>2 \\ \\ \\3x+4>22 \\ \\3x>22-4\\ \\3x>18\ \ /:3\\ \\x>6
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What is 76 percent of 33
gulaghasi [49]

Answer:

25.08

Step-by-step explanation:

do 76 ÷ 100, and then multiply that by 33

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3 years ago
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Plz answer all measures plz ASAP thx
LiRa [457]

Lines <em>a</em> and <em>b</em> are parallel, so lines <em>p</em>, <em>q</em>, and <em>t</em> are considered to be transversals. To solve this, you make use of the fact that alternate interior angles are equal, as are alternate exterior angles, as are corresponding angles. Of course any linear pair of angles is supplementary.

∠1 = 90° — corresponding angle to the right angle above it

∠2 = 68° — the sum of 22° and angles 1 and 2 is 180°

∠3 = 112° — supplementary to angle 2 (and the sum of 22° and 90°, opposite interior angles of the triangle)

∠4 = 112° — equal to angle 3

∠5 = 68° — equal to angle 2; supplementary to angle 4

∠6 = 56° — base angle of isosceles triangle with 68° at the apex; the complement of half that apex angle

∠7 = 124° — supplementary to the other base angle, which is equal to angle 6; also the sum of angles 5 and 6

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3 0
3 years ago
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Fission tracks are trails found in uranium-bearing minerals, left by fragments released during fission events. An article report
Harlamova29_29 [7]

Answer:

Mean track length for this rock specimen is between 10.463 and 13.537

Step-by-step explanation:

99% confidence interval for the mean track length for rock specimen can be calculated using the formula:

M±\frac{t*s}{\sqrt{N}} where

  • M is the average track length (12 μm) in the report
  • t is the two tailed t-score in 99% confidence interval (2.977)
  • s is the standard deviation of track lengths in the report (2 μm)
  • N is the total number of tracks (15)

putting these numbers in the formula, we get confidence interval in 99% confidence as:

12±\frac{2.977*2}{\sqrt{15}} =12±1.537

Therefore, mean track length for this rock specimen is between 10.463 and 13.537

4 0
3 years ago
I REALLY NEED HELP
garri49 [273]
Look at the picture.

1.
y=mx+b\\\\m=\dfrac{9}{3}=3\\\\b=4\\\\\boxed{f(x)=3x+4}

2.
f(x)=-2x+3\\\\for\ x=-2\to f(-2)=-2\cdot(-2)+3=4+3=7\to(-2;\ 7)\\for\ x=-1\to f(-1)=-2\cdot(-1)+3=2+3=5\to(-1;\ 5)\\for\ x=0\to f(0)=-2\cdot0+3=3\to(0;\ 3)\\for\ x=1\to f(1)=-2\cdot1+3=-2+3=1\to(1;\ 1)

3.
y=mx+b\\\\m=\dfrac{3}{6}=\dfrac{1}{2}=0.5\\\\b=-3\\\\\boxed{f(x)=0.5x-3}
for\ x=-6\to f(-6)=-6\\for\ x=-2\to f(-2)=-4\\for\ x=0\to f(0)=-3\\for\ x=6\to f(6)=0

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3 years ago
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FrozenT [24]

Answer:

Step-by-step explanation:

I will help u if you help me????

7 0
2 years ago
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