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Lina20 [59]
3 years ago
10

Use the triangles above. Find the side lengths and perimeters if x=5.

Mathematics
1 answer:
NeTakaya3 years ago
3 0

First triangle side lengths: 9, 4, 8

First triangle perimeter: 21

Second triangle side lengths: 18, 8, 16

Second triangle perimeter: 42

Please consider brainliest!

✉️ If any further questions, inbox me! ✉️

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Two angle measure of a quadrilateral 90 and 90. What could the measure of the other two angle be?
kompoz [17]
So, I think that your answer will be D because it has to end at a degree of 180
7 0
4 years ago
How do I answer this dot plot?
Anna11 [10]
It would seem the problem requires you compute the mean, median, and standard deviation of each of the data sets. This is nicely done using the statistics functions of your graphing calculator.

Data Set A (mean, median, standard deviation) = (29.7, 29.5, √12.71)
Data Set B (mean, median, standard deviation) = (27.95, 27, √6.1475)

1. False. Data Set B is skewed to the right.
2. TRUE. 27.95 is within 3 of 29.7.
3. False. √12.71 ≠ √6.1475
4. False. The mean of Data Set A is 29.7.
5. False. Data Set A has the higher standard deviation.
6. TRUE. 29.7 is close to 29.5.

The true statements are
• the 2nd one
• the 6th one
3 0
3 years ago
Evaluate ∫SF⃗ ⋅dA⃗ , where F⃗ =(bx/a)i⃗ +(ay/b)j⃗ and S is the elliptic cylinder oriented away from the z-axis, and given by x2/
Norma-Jean [14]

Answer:

Therefore surface integral is \pi(a^2+b^2)c-0-0=\pi(a^2+b^2)c.

Step-by-step explanation:

Given function is,

\vec{F}=\frac{bx}{a}\uvec{i}+\frac{ay}{b}\uvec{j}

To find,

\int\int_{S}\vec{F}dS  

where S=A=surfece of elliptic cylinder we have to apply Divergence theorem so that,

\int\int_{S}\vec{F}dS

=\int\int\int_V\nabla.\vec{F}dV

=\int\int\int_V(\frac{b}{a}+\frac{a}{b})dV  

=\frac{a^2+b^2}{ab}\int\int\int_VdV

=\frac{a^2+b^2}{ab}\times \textit{Volume of the elliptic cylinder}

=\frac{a^2+b^2}{ab}\times \pi ab\times 2c=\pi (a^2+b^2)c

  • If unit vector \cap{n} directed in positive (outward) direction then z=c and,

\int\int_{S_1}\vex{F}.dS_1=\int\int_{S_1} . dA      

=\int\int_{S_1}.dA=0

  • If unit vector \cap{n} directed in negative (inward) direction then z=-c and,

\int\int_{S_2}\vex{F}.dS_2=\int\int_{S_2}. -dA      

=\int\int_{S_2}. -dA=0

Therefore surface integral without unit vector of the surface is,

\pi(a^2+b^2)c-0-0=\pi(a^2+b^2)c

5 0
4 years ago
A rocket fired straight up is tracked by an observer from the ground a mile away. (a) show that when the angle of elevation is t
Dahasolnce [82]
Let the height of the rocket be x, then
tan theta = h / 1 mile
h = 1 tan theta miles

But 1 mile = 5280 feets

Therefore, the height in feet = 5280 tan theta.
4 0
3 years ago
PLEASE HELP!!
Bezzdna [24]

Answer:

y=-\dfrac{9}{40}x^2+\dfrac{441}{40}\\ \\y=-\dfrac{9}{160}x^2+\dfrac{1,521}{160}

Step-by-step explanation:

If a parabola has its vertex on the y-axis, then its equation is

y=ax^2+b

This parabola passes through the point R(3,9), then

9=a\cdot 3^2+b\\ \\9=9a+b

The area of the right triangle PQR is

A_{PQR}=\dfrac{1}{2}\cdot PQ\cdot QR

Find PQ and QR, if P(x_1,0),\ Q(3,0),\ R(3,9):

PQ=\sqrt{(x_1-3)^2+(0-0)^2}=|x_1-3|\\\\QR=\sqrt{(3-3)^2+(0-9)^2}=9

Now,

40.5=\dfrac{1}{2}\cdot |x_1-3|\cdot 9\\ \\90=9|x_1-3|\\ \\|x_1-3|=10\\ \\x_1-3=10\ \text{or}\ x_1-3=-10\\ \\x_1=13\ \text{or }x_1=-7

We get two possible points P_1(-7,0) and P_2(13,0).

For point P_1:\\

0=a\cdot (-7)^2+b\\ \\49a+b=0

So,

b=-49a\\ \\9=9a-49a\\ \\-40a=9\\ \\a=-\dfrac{9}{40}\\ \\b=\dfrac{441}{40}\\ \\y=-\dfrac{9}{40}x^2+\dfrac{441}{40}

For point P_2:\\

0=a\cdot (13)^2+b\\ \\169a+b=0

So,

b=-169a\\ \\9=9a-169a\\ \\-160a=9\\ \\a=-\dfrac{9}{160}\\ \\b=\dfrac{1,521}{160}\\ \\y=-\dfrac{9}{160}x^2+\dfrac{1,521}{160}

5 0
3 years ago
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