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pychu [463]
2 years ago
11

Anyone knows the answer?????????????

Mathematics
2 answers:
Softa [21]2 years ago
4 0

Answer:

noooooooooooiiiiiiooooooooooooooooo

andrey2020 [161]2 years ago
3 0

Answer:

No I don't know if they also know then they will tell don't know that then how to tell

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What is data? what can when can we use data?
sukhopar [10]
Data can be on a graph like if the weather increase 2 degrees
3 0
3 years ago
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Find the volume of the solid under the plane 5x + 9y − z = 0 and above the region bounded by y = x and y = x4.
svp [43]
<span>For the plane, we have z = 5x + 9y

For the region, we first find its boundary curves' points of intersection.
x = x^4 ==> x = 0, 1.

Since x > x^4 for y in [0, 1],

The volume of the solid equals

\int\limits^1_0 { \int\limits_{x^4}^x {(5x+9y)} \, dy } \, dx = \int\limits^1_0 {\left[5xy+ \frac{9}{2} y^2\right]_{x^4}^{x}} \, dx  \\  \\ =\int\limits^1_0 {\left[\left(5x(x)+ \frac{9}{2} (x)^2\right)-\left(5x(x^4)+ \frac{9}{2} (x^4)^2\right)\right]} \, dx  \\  \\ =\int\limits^1_0 {\left(5x^2+ \frac{9}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx =\int\limits^1_0 {\left( \frac{19}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx \\  \\ =\left[ \frac{19}{6} x^3- \frac{5}{6} x^6- \frac{1}{2} x^9\right]^1_0

=\frac{19}{6} - \frac{5}{6} - \frac{1}{2} =\bold{ \frac{11}{6} \ cubic \ units}</span>
8 0
3 years ago
At a local Brownsville play production, 420 tickets were sold. The ticket
Dahasolnce [82]

Answer:

jsdcjdvnjkdnjnjdanskcbanknqnjfkrbgiyrwhgondfkv

Step-by-step explanation:

8 0
3 years ago
Find the product of (3^6)(3)(3^5) Express your answer using exponents
VikaD [51]

Answer:

3^12

Step-by-step explanation:

When multiplying the same number with different exponents you have to add the the exponents. For example:

a^x + a^y = a ^{x+y}

So (3^6)(3)(3^5) or (3^6)(3^1)(3^5) = 3^(6+1+5) = 3^12

6 0
3 years ago
Read 2 more answers
6. Two observers, 7220 feet apart, observe a balloonist flying overhead between them. Their measures of the
MaRussiya [10]

Answer:

The ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

Step-by-step explanation:

Let's call:

h the height of the ballonist above the ground,

a the distance between the two observers,

a_1 the horizontal distance between the first observer and the ballonist

a_2 the horizontal distance between the second observer and the ballonist

\alpha _1 and \alpha _2 the angles of elevation meassured by each observer

S the area of the triangle formed with the observers and the ballonist

So, the area of a triangle is the length of its base times its height.

S=a*h (equation 1)

but we can divide the triangle in two right triangles using the height line. So the total area will be equal to the addition of each individual area.

S=S_1+S_2 (equation 2)

S_1=a_1*h

But we can write S_1 in terms of \alpha _1, like this:

\tan(\alpha _1)=\frac{h}{a_1} \\a_1=\frac{h}{\tan(\alpha _1)} \\S_1=\frac{h^{2} }{\tan(\alpha _1)}

And for S_2 will be the same:

S_2=\frac{h^{2} }{\tan(\alpha _2)}

Replacing in the equation 2:

S=\frac{h^{2} }{\tan(\alpha _1)}+\frac{h^{2} }{\tan(\alpha _2)}\\S=h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})

And replacing in the equation 1:

h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})=a*h\\h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}

So, we can replace all the known data in the last equation:

h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}\\h=\frac{7220 ft}{(\frac{1 }{\tan(35.6)}+\frac{1}{\tan(58.2)})}\\h=3579,91 ft

Then, the ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

6 0
3 years ago
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