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Anna71 [15]
3 years ago
11

Pat thinks of a number that is greater than 4 and less than 5.He doubles the number

Mathematics
1 answer:
soldi70 [24.7K]3 years ago
4 0

Answer:

If you know your doubles, then it should be easy to go from 4 + 4 = 8 to 4 + 5 = 9.

Step-by-step explanation:

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Write in point slope form an equation of the line that passes through the given point and has the given slope: (0,1); m= 2
sergey [27]

           - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

\diamond\large\blue\textsf{\textbf{\underline{\underline{Question:-}}}}\diamond

         Write an equation for the line that passes through (0, 1) and has a slope of 2 (in point-slope form).

\diamond\large\blue\textsf{\textbf{\underline{\underline{Answer and How to Solve:-}}}}\diamond

<u>Point-slope form</u>:-

      \sf{y-y_1=m(x-x_1)}

Substitute 1 for y₁, 2 for m, and 0 for x₁:-

\sf{y-1=2(x-0)}

So we conclude that Option B is correct.

<h3>Good luck.</h3>

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3 0
2 years ago
7.7.33
GenaCL600 [577]
We know that

At sea level, the height is 0 and
the pressure is 98 kilopascals

At 1000 ft, the height is 1000 and
the pressure <span>decreases about 11.41%

</span>i(100%-11.41%)/100----> (0.8859)

therefore
1) at 1000 ft--------> the pressure is 98*(0.8859)----------> 86.82 kilopascals
2) at 2000 ft--------> the pressure is 86.82*(0.8859)-------> 76.91 kilopascals
3) at 3000 ft--------> the pressure is 76.91*(0.8859)-------> 68.14 kilopascals
4) at 4000 ft---------> the pressure is 68.14*(0.8859)-------> 60.36 kilopascals


the answer is
<span>The pressure at an altitude of 4000 m is about </span>60.36 kilopascals
7 0
3 years ago
Line passes through point (1, 7) and point (-2, 4). What is the slope of this line?
Drupady [299]

The slope of the line is 1.

The slope of the line is 1 because the line moves ror goes up from left to right, making it a positive. The line also moves one block over and one block up.

3 0
3 years ago
Please I need x and y
Kryger [21]
It could be anything they are both variables.
3 0
3 years ago
Y = 3 sin2x, y = 0, 0 ≤ x ≤ π; about the x−axis
Dominik [7]
I assume you're revolving the region with those bounds about the x-axis, and supposed to find the volume.

Via the disk method,

\displaystyle\pi\int_0^\pi(3\sin2x)^2\,\mathrm dx=9\pi\int_0^\pi\sin^22x\,\mathrm dx

Recall the half-angle identity for sine:

\sin^2t=\dfrac{1-\cos2t}2
\implies\displaystyle\frac{9\pi}2\int_0^\pi(1-\cos4x)\,\mathrm dx
=\displaystyle\frac{9\pi}2\left(x-\frac14\sin4x\right)\bigg|_{x=0}^{x=\pi}
=\dfrac{9\pi^2}2
8 0
3 years ago
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