Find the value of x.

A section in a stained glass window is shaped like a trapezoid. The top base is 2 centimeters and the bottom base is 4.5 centime

Elza [17]

Answer:

2.8 cm

Step-by-step explanation:

The area of a trapezoid is calculated using the formula:

$A=\frac{(B+b)h}{2}$

where

B is the length of the major base

b is the length of the minor base

h is the height of the trapezoid

In this problem, the glass window has the shape of a trapezoid. We know that:

B = 2 cm is the length of the top base

b = 4.5 cm is the length of the bottom base

$A=9.1 cm^2$ is the area of the trapezoid

Solving for h, we can find the height of the stained glass window:

$h=\frac{2A}{B+b}=\frac{2(9.1)}{2+4.5}=2.8 cm$

8 0

3 years ago

Expand(2x-1)(x+5) please answer this ASAP thank you.

hjlf

Answer:

2x² + 9x - 5

Step-by-step explanation:

Given

(2x - 1)(x + 5)

Each term in the second factor is multiplied by each term in the first factor, that is

2x(x + 5) - 1(x + 5) ← distribute both parenthesis

= 2x² + 10x - x - 5 ← collect like terms

= 2x² + 9x - 5

4 0

3 years ago

Evaluate the surface integral. s y ds, s is the helicoid with vector equation r(u, v) = u cos(v), u sin(v), v , 0 ≤ u ≤ 6, 0 ≤ v

Juliette [100K]

Compute the surface element:

$\mathrm dS=\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv$

$\vec r(u,v)=(u\cos v,u\sin v,v)\implies\begin{cases}\vec r_u=(\cos v,\sin v,0)\\\vec r_v=(-u\sin v,u\cos v,1)\end{cases}$

$\|\vec r_u\times\vec r_v\|=\sqrt{\sin^2v+(-\cos v)^2+u^2}=\sqrt{1+u^2}$

So the integral is

$\displaystyle\iint_Sy\,\mathrm dS=\int_0^\pi\int_0^6u\sin v\sqrt{1+u^2}\,\mathrm du\,\mathrm dv$

$=\displaystyle\left(\int_0^\pi\sin v\,\mathrm dv\right)\left(\int_0^6u\sqrt{1+u^2}\,\mathrm du\right)$

$=\dfrac23(37^{3/2}-1)$

4 0

3 years ago

If vector u has its initial point at (-7, 3) and its terminal point at (5, -6), u =

attashe74 [19]

First of all, let θθ be some angle in (0,π)(0,π). Then

θθ is acute ⟺⟺ θ<π2θ<π2 ⟺⟺ cosθ>0cos⁡θ>0.θθ is right ⟺⟺ θ=π2θ=π2 ⟺⟺ cosθ=0cos⁡θ=0.θθ is obtuse ⟺⟺ θ>π2θ>π2 ⟺⟺ cosθ<0cos⁡θ<0.

Now, to see if (say) angle AA of the triangle ABCABC is acute/right/obtuse, we need to check whether cos∠BACcos⁡∠BAC is positive/zero/negative. But what is cos∠BACcos⁡∠BAC? It is the angle made by the vectors AB−→−AB→ and AC−→−AC→. (When you are computing the angle at a particular vertex vv, you should make sure that both the vectors corresponding to the two adjacent sides have that vertex vv as the initial point.) We will first compute these two vectors:

AB−→−=(0,0,0)−(1,2,0)=(−1,−2,0)AB→=(0,0,0)−(1,2,0)=(−1,−2,0)AC−→−=(−2,1,0)−(1,2,0)=(−3,−1,0)AC→=(−2,1,0)−(1,2,0)=(−3,−1,0)Therefore, the angle between these vectors is given by:cos∠BAC=AB−→−⋅AC−→−|AB−→−||AC−→−|=…(1)(1)cos⁡∠BAC=AB→⋅AC→|AB→||AC→|=…Can you take it from here? From the sign of this value, you should be able to decide if angle AA is acute/right/obtuse.

Now, do the same procedure for the remaining two angles BB and CC as well. That should help you solve the problem.

A shortcut. Since you are not interested in the actual values of the angles, but you need only whether they are acute, obtuse or right, it is enough to compute only the sign of the numerator (the dot product between the vectors) in formula (1). The denominator is always positive.

6 0

3 years ago

What is the volume of a sphere with a diameter of 9.1 m, rounded to the nearest tenth of a cubic meter?

muminat

Hello the answer would be approximately 394.57 good luck with the test

7 0

2 years ago