What is the coefficient of the second term of the expressions 3x⁴ + x²?

You're given two side lengths of 3 centimeters and 5 centimeters. Which measurement can you use for the length of the third side

igomit [66]

Answer: To find the third side of a triangle (there are two answers you could get), you need to use the Pythagorean formula: a^2+b^2=c^2. First answer, 3^2+5^2=c^2. Solve for c. c=5.83. The second answer, 3^2+b^2=5^2. Solve for b. b=4.

\

5 0

3 years ago

The surface area of a cone with radius r units and slant height s units is shown below:

Ganezh [65]

Surface area of a cone = π{rs+r²} =πr{s+r} part 1:if π is 6 units and s is 4 units then r would be 3.14 ×6 {4+6} part 2: area of a cone = 3.14 ×6×10 =31.4×6 = 188.4 units² the surface area = 188.4 units² hope i helped

4 0

3 years ago

Read 2 more answers

Find the area of the region that lies inside the first curve and outside the second curve.

marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = $\dfrac{1}{2}$

$\theta = -\dfrac{\pi}{3}, \ \ \dfrac{\pi}{3}$

Now, the area of the region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \ \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ 5^2 d \theta$

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \ \theta d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ d \theta$

$A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} \dfrac{cos \ 2 \theta +1}{2} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}$

$A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} {cos \ 2 \theta +1} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}} \ \ - \dfrac{25}{2} \begin {bmatrix} \dfrac{2 \pi}{3} \end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3}) \end {bmatrix} - \dfrac{25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} + \dfrac{\pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}$

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx

7 0

3 years ago

B) i) Solve - 4(3+x) + 5x + 3

Nina [5.8K]

Answer:

9x+15

Step-by-step explanation:

7 0

3 years ago

A spinner is spun once ( The spinner has 8 sections in it) What is the probability that the spinner will land on 5 or 7? A.8% B.

nordsb [41]

D. 25% as there are 8 sections and two wanted outcomes this can be simplified into a 1/4 so 1/4=25%

3 0

3 years ago