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Harlamova29_29 [7]
3 years ago
14

What is the reduce form of 10/ab+x

Mathematics
1 answer:
madam [21]3 years ago
7 0

1.

\frac{10}{2 {ab}^{2} }  \times   \frac{ {a}^{2}  {b}^{2}}{5}

=  \frac{5}{2a {b}^{2} }  \times   {a}^{2}  {b}^{2}

=  \frac{5}{2a}  \times  {a}^{2}

=  \frac{5 {a}^{2} }{2a}

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A magician performing a magic act has asked for 2 volunteers from the audience to help with his routine. If there are 250 people
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The answer is 125. If you were to pair 250 people in 2, that would be 250 divided by 2. And 250 divided by 2 is 125.
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Question is in the picture:)
Bingel [31]

Answer: It would probably be 30

Step-by-step explanation:

The ratio is 10:0 , 20:10 , 30:20 , 40:30 , 50:40, 60:50 , etc…

so basically you just need to know the ratios!

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RideAnS [48]

Answer:

The equation of the regression line for the following data is:

y=2.4x+120.1

Step-by-step explanation:

We are given a set of data values in table form as:

           Day(x)                  Number of visitors(y)

               1                              120

               2                             124

               3                             130

               4                              131

               5                              135

               6                              132

               7                              135

Hence, when we draw a scatter plot with the help of these data points using the linear regression calculator we see that the regression line is a line with y-intercept as (0,120.1) and x-intercept as  (-50.04,0)

Hence, the regression line for the following data is:

                   y=2.4x+120.1

8 0
3 years ago
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Evaluate lim x→∞ (3x+1)^(4/x), using l'hospital's rule as needed. show all work using proper notation. as you show your work, if
Alborosie
\displaystyle\lim_{x\to\inty}(3x+1)^{4/x}=\lim_{x\to\infty}e^{\ln(3x+1)^{4/x}}=e^{\lim\limits_{x\to\infty}\ln(3x+1)^{4/x}}

\displaystyle\lim_{x\to\infty}\ln(3x+1)^{4/x}=\lim_{x\to\infty}\frac{4\ln(3x+1)}x\stackrel{\mathrm{LHR}}=\lim_{x\to\infty}\frac{4\frac3{3x+1}}1=\lim_{x\to\infty}\frac{12}{3x+1}=0

\implies\displaystyle\lim_{x\to\infty}(3x+1)^{4/x}=e^0=1
4 0
4 years ago
Select all the conditions for which it is possible to construct a triangle. (7.G.1.2) Group of answer choices a. A triangle with
saw5 [17]

Answer:

  b, d, e, f

Step-by-step explanation:

Here are the applicable restrictions:

  • The sum of angles in a triangle is 180°, no more, no less.
  • The sum of the lengths of the two shortest sides exceeds the longest side.
  • When two sides and the angle opposite the shortest is given, the sine of the given angle must be at most the ratio of the shortest to longest sides.

a. A triangle with angle measures 60°, 80°, and 80° (angle sum ≠ 180°, not OK)

b. A triangle with side lengths 4 cm, 5 cm, and 6 cm (4+5 > 6, OK)

c. A triangle with side lengths 4 cm, 5 cm, and 15 cm (4+5 < 15, not OK)

d. A triangle with side lengths 4 cm, 5 cm, and a 50° angle across from the 4 cm side (sin(50°) ≈ 0.766 < 4/5, OK)

e. A triangle with angle measures 30° and 60°, and an included 3 cm side length (OK)

f. A triangle with angle measures 60°, 20°, and 100° (angle sum = 180°, OK)

_____

<em>Additional comment</em>

In choice "e", two angles and the side between them are specified. As long as the sum of the two angles is less than 180°, a triangle can be formed. The length of the side is immaterial with respect to whether a triangle can be made.

__

The congruence postulates for triangles are ...

  SSS, SAS, ASA, AAS, and HL

These essentially tell you the side and angle specifications necessary to define <em>a singular triangle</em>. As we discussed above, the triangle inequality puts limits on the side lengths specified in SSS. The angle sum theorem puts limits on the angles when only two are specified (ASA, AAS).

In terms of sides and angles, the HL postulate is equivalent to an SSA theorem, where the angle is 90°. In that case, the angle is opposite the longest side (H). In general, SSA will specify a singular triangle when the angle is opposite the <em>longest</em> specified side, regardless of that angle's measure. However, when the angle is opposite the <em>shortest</em> specified side, the above-described ratio restriction holds. If the sine of the angle is <em>less than</em> the ratio of sides, then <em>two possible triangles are specified</em>.

4 0
2 years ago
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