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4 years ago

Evaluate the expression when a = 2 and b = 4. 6a – 2b + 5 A. 43 B. 1 C. 45 D. 9

Mathematics

1 answer:

tamaranim1 [39]4 years ago

6 0

Hi there!
6(2) -2(4) +5
12-8+5
So D would be correct.

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Please help with this geometry question

Andrew [12]

It has been proven that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

<h3>How to prove a Line Segment?</h3>

We know that in a triangle if one angle is 90 degrees, then the other angles have to be acute.

Let us take a line l and from point P as shown in the attached file, that is, not on line l, draw two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle.

In ΔPNM, ∠N = 90°

∠P + ∠N + ∠M = 180° (Angle sum property of a triangle)

∠P + ∠M = 90°

Clearly, ∠M is an acute angle.

Thus; ∠M < ∠N

PN < PM (The side opposite to the smaller angle is smaller)

Similarly, by drawing different line segments from P to l, it can be proved that PN is smaller in comparison to all of them. Therefore, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Read more about Line segment at; brainly.com/question/2437195

#SPJ1

7 0

2 years ago

Use the upper and lower sums to approximate the area of the region using the given number of subintervals (of equal width)

Mashutka [201]

From the figure shown, the interval is divided into 5 equal parts making each subinterval to be 0.2.

Part A:

$y= \sqrt{1-x^2}$

The approximate the area of the region shown in the figure using the lower sums is given by:

$Area= [y(0.2)\times0.2]+[y(0.4)\times0.2]+[y(0.6)\times0.2]+[y(0.8)\times0.2] \\ +[y(1)\times0.2] \\ \\ =[\sqrt{1-(0.2)^2}\times0.2]+[\sqrt{1-(0.4)^2}\times0.2]+[\sqrt{1-(0.6)^2}\times0.2] \\ +[\sqrt{1-(0.8)^2}\times0.2]+[\sqrt{1-(1)^2}\times0.2] \\ \\ =(0.9798\times0.2)+(0.9165\times0.2)+(0.8\times0.2)+(0.6\times0.2)+(0\times0.2) \\ \\ =0.196+0.183+0.16+0.12=0.659$

Part B:

The approximate the area of the region shown in the figure using the lower sums is given by:

$Area= [y(0)\times0.2]+[y(0.2)\times0.2]+[y(0.4)\times0.2]+[y(0.6)\times0.2] \\ +[y(0.8)\times0.2] \\ \\ =[\sqrt{1-(0)^2}\times0.2]+[\sqrt{1-(0.2)^2}\times0.2]+[\sqrt{1-(0.4)^2}\times0.2] \\ +[\sqrt{1-(0.6)^2}\times0.2] +[\sqrt{1-(0.8)^2}\times0.2] \\ \\ =(1\times0.2)+(0.9798\times0.2)+(0.9165\times0.2)+(0.8\times0.2)+(0.6\times0.2) \\ \\ =0.2+0.196+0.183+0.16+0.12=0.859$

Part C:

The approximate area of the given region is given by

$Area= \frac{0.659+0.859}{2} = \frac{1.518}{2} =0.759$

7 0

4 years ago

Abdul is creating a cushion that is 1 foot by 2 feet by 3 feet. He wants to find the least amount of fabric he will need to cove

anygoal [31]

Answer:

22 ft²

Step-by-step explanation:

he needs 2 sides that are 2 x 3 feet

2 sides that are 3 x 1 and 2 sides that are 2 x 1

add these up to get 22 ft

7 0

3 years ago

Read 2 more answers

Find the area of the surface. the part of the paraboloid z=4-x^2-y^2 that lies above the xy-plane

11Alexandr11 [23.1K]

Parameterize the surface (call it $\mathcal S$ ) by

$\mathbf s(u,v)=\langle u\cos v,u\sin v,4-u^2\rangle$

with $0\le u\le2$ and $0\le v\le2\pi$ . Then the surface element is

$\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|=u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv$

The area of $\mathcal S$ is then given by the surface integral

$\displaystyle\iint_{\mathcal S}\mathrm dS=\int_{v=0}^{v=2\pi}\int_{u=0}^{u=2}u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv$
$=\displaystyle2\pi\int_{u=0}^{u=2}u\sqrt{1+4u^2}\,\mathrm du=\dfrac{(17^{3/2}-1)\pi}6\approx36.1769$

6 0

3 years ago

The graph shows the square root of the parent function. which statement is true?

Nady [450]

B is the answer for this problem

4 0

3 years ago