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3 years ago

Need help with the answer ASAP! No links by the way!

Mathematics

1 answer:

zvonat [6]3 years ago

4 0

Answer:

Hewo Asuna here

There’s no link sorry

Step-by-step explanation:

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Please help me with this question

adoni [48]

Answer:

$y = e^{x^2 - 3x}\\\\\\Let \ u = x^2 - 3x\\\\y = e^u\\\\\frac{dy}{du} = e^u\\\\\frac{du}{dx} = 2x - 3\\\\y' = \frac{dy}{dx} = \frac{dy}{du} *\frac{du}{dx} = e^u * (2x -3)\\\\Substitute \ u = x^2 -3\\\\y' = (2x -3)e^{x^2-3x}\\\\option A$

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3 years ago

Brady made a scale drawing of a rectangular swimming pool on a coordinate grid. The points (-20, 25), (30, 25), (30, -10) and (-

djverab [1.8K]

Answer:

Length = 50 units

width = 35 units

Step-by-step explanation:

Let A, B, C and D be the corner of the pools.

Given:

The points of the corners are.

$A(x_{1}, y_{1}})=(-20, 25)$

$B(x_{2}, y_{2}})=(30, 25)$

$C(x_{3}, y_{3}})=(30, -10)$

$D(x_{4}, y_{4}})=(-20, -10)$

We need to find the dimension of the pools.

Solution:

Using distance formula of the two points.

$d(A,B)=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ ----------(1)

For point AB

Substitute points A(30, 25) and B(30, 25) in above equation.

$AB=\sqrt{(30-(-20))^{2}+(25-25)^{2}}$

$AB=\sqrt{(30+20)^{2}}$

$AB=\sqrt{(50)^{2}$

AB = 50 units

Similarly for point BC

Substitute points B(-20, 25) and C(30, -10) in equation 1.

$d(B,C)=\sqrt{(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}$

$BC=\sqrt{(30-30)^{2}+((-10)-25)^{2}}$

$BC=\sqrt{(-35)^{2}}$

BC = 35 units

Similarly for point DC

Substitute points D(-20, -10) and C(30, -10) in equation 1.

$d(D,C)=\sqrt{(x_{3}-x_{4})^{2}+(y_{3}-y_{4})^{2}}$

$DC=\sqrt{(30-(-20))^{2}+(-10-(-10))^{2}}$

$DC=\sqrt{(30+20)^{2}}$

$DC=\sqrt{(50)^{2}}$

DC = 50 units

Similarly for segment AD

Substitute points A(-20, 25) and D(-20, -10) in equation 1.

$d(A,D)=\sqrt{(x_{4}-x_{1})^{2}+(y_{4}-y_{1})^{2}}$

$AD=\sqrt{(-20-(-20))^{2}+(-10-25)^{2}}$

$AD=\sqrt{(-20+20)^{2}+(-35)^{2}}$

$AD=\sqrt{(-35)^{2}}$

AD = 35 units

Therefore, the dimension of the rectangular swimming pool are.

Length = 50 units

width = 35 units

7 0

3 years ago

Divide for 2class .45÷6789

FrozenT [24]

0.00006628369 is your answer for 0.45/6789

8 0

3 years ago

Find the missing angle measures. What is m∠2? a 40 b 60 c 50 d 100

erica [24]

The Answer is C.50
Sixty

6 0

2 years ago

Read 2 more answers

Find the point(s) of intersection (if any) of the plane and the line. Also, determine whether the line lies in the plane. 2x - 2

s2008m [1.1K]

Answer:

the intersection is the point P=(x,y,z)=(8,9,14) and the line does not lie in the plane

Step-by-step explanation:

from the equation of the line

x - 1/2 = y + (3/2)/-1 = (z + 1) / 2 = t (parameter)

then the parametric equation of the line is

x= 1/2 +t

y = (3/2) + t

z = (-1) + 2*t

therefore from the equation of the plane

2x - 2y + z = 12

2*(1/2 +t) - 2[(3/2) + t] + [(-1) + 2*t] = 12

1+2*t - 3 -2*t -1 + 2*t = 12

-3 + 2*t = 12

t= 15/2

therefore there is only one intersection of the line with the plane ( then the line does not lie in the plane , since there would be infinite intersection points). The intersection is

x= 1/2 +t = 1/2 +15/2 = 8

y = (3/2) + t = (3/2) + 15/2 = 9

z = (-1) + 2*t = (-1) + 2*15/2 = 14

thus the intersection point P=(x,y,z)=(8,9,14)

7 0

2 years ago