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NARA [144]
3 years ago
15

Help me please and thank you

Mathematics
2 answers:
Anuta_ua [19.1K]3 years ago
5 0
The slope is 3 y-intercept is 7 and the equation would be y=3x+7
ivann1987 [24]3 years ago
3 0
Slope-3
Y-intercept- 7

y=3x+7
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Determine which characteristics are true of vertical angles
liq [111]

Answer:

1, 2 and 4 option - Yes

3 option - No

Step-by-step explanation:

<u>Definition: </u>Vertical angles are the angles opposite each other when two lines cross.

By the definition options

  • they are formed by intersecting lines,
  • they are opposite to each other

are true.

Vertical Angles Theorem states that vertical angles are congruent. Thus, option

  • they are congruent

is true.

Option

  • they are always supplementary

is false, because supplementary are angles that add up to 180°, but vertical angles can have different measures which do not add up to 180°.

7 0
4 years ago
Read 2 more answers
Amie works 6 hours each Saturday and 4 hours each Monday. She also makes $20 a week as a dog walker. She makes a total of $130 e
vivado [14]

Answer:

11 per hour

Step-by-step explanation:

130 - 20=110/(6+4)=11

3 0
4 years ago
Read 2 more answers
Find the Volumes?<br> I need help with both of these, please!!!
Arada [10]

Answer:

1. 36

2. 5390

Step-by-step explanation:

1. V=1wh/3

4 x 3 x 9 = 108

108/3 = 36

2. V=lwh/3

22 x 21 x 35 = 16170

16170/3 = 5390

4 0
3 years ago
Read 2 more answers
How do i graph the parabolas y = 1/32x^2
Pachacha [2.7K]
y=\frac{1}{32}x^2 \\ \\x=-16 \Rightarrow  y=\frac{1}{32} *(-16)^2=\frac{1}{\not32^1}*\not256^8=8\\ \\x=-8 \Rightarrow  y=\frac{1}{32} *(-8)^2=\frac{1}{\not32^1}*\not64^2=2\\ \\x=0 \Rightarrow  y=\frac{1}{32} *(0)^2=\frac{1}{\not32^1}*0=0 \\ \\x=8 \Rightarrow  y=\frac{1}{32} *(8)^2=\frac{1}{\not32^1}*\not64^2=2 \\ \\x= 16 \Rightarrow  y=\frac{1}{32} *( 16)^2=\frac{1}{\not32^1}*\not256^8=8

7 0
3 years ago
Let R be the region in the first quadrant of the​ xy-plane bounded by the hyperbolas xyequals​1, xyequals9​, and the lines yequa
Tema [17]

Answer:

The area can be written as

\int\limits_1^2 \int\limits_1^3 u(\frac{1}{v} - v \, ln(v)) \, du \, dv = 0.2274

And the value of it is approximately 1.8117

Step-by-step explanation:

x = u/v

y = uv

Lets analyze the lines bordering R replacing x and y by their respective expressions with u and v.

  • x*y = u/v * uv = u², therefore, x*y = 1 when u² = 1. Also x*y = 9 if and only if u² = 9
  • x=y only if u/v = uv, And that only holds if u = 0 or 1/v = v, and 1/v = v if and only if v² = 1. Similarly y = 4x if and only if 4u/v = uv if and only if v² = 4

Therefore, u² should range between 1 and 9 and v² ranges between 1 and 4. This means that u is between 1 and 3 and v is between 1 and 2 (we are not taking negative values).

Lets compute the partial derivates of x and y over u and v

x_u = 1/v

x_v = u*ln(v)

y_u = v

y_v = u

Therefore, the Jacobian matrix is

\left[\begin{array}{ccc}\frac{1}{v}&u \, ln(v)\\v&u\end{array}\right]

and its determinant is u/v - uv * ln(v) = u * (1/v - v ln(v))

In order to compute the integral, we can find primitives for u and (1/v-v ln(v)) (which can be separated in 1/v and -vln(v) ). For u it is u²/2. For 1/v it is ln(v), and for -vln(v) , we can solve it by using integration by parts:

\int -v \, ln(v) \, dv = - (\frac{v^2 \, ln(v)}{2} - \int \frac{v^2}{2v} \, dv) = \frac{v^2}{4} - \frac{v^2 \, ln(v)}{2}

Therefore,

\int\limits_1^2 \int\limits_1^3 u(\frac{1}{v} - v \, ln(v)) \, du \, dv = \int\limits_1^2 (\frac{1}{v} - v \, ln(v) ) (\frac{u^2}{2}\, |_{u=1}^{u=3}) \, dv= \\4* \int\limits_1^2 (\frac{1}{v} - v\,ln(v)) \, dv = 4*(ln(v) + \frac{v^2}{4} - \frac{v^2\,ln(v)}{2} \, |_{v=1}^{v=2}) = 0.2274

4 0
4 years ago
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