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2 years ago

7

the formula for the area of a triangle is 1/2bh=A. Which of the not a first step to rearrange this formula to find to find the b

ase of the triangke?

1 answer:

Travka [436]2 years ago

5 0

Answer:

base=area=1/2b×h

Step-by-step explanation:

it's a pleasure to help you goodbye. if I'm wrong tell me right away.

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THE HCF OF TWO NUMBERS 2952 AND 2256 IS 24 FIND THEIR LCM<br>

MrRissso [65]

Step-by-step explanation:

by which method defination method, prime factorization or division method

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2 years ago

To the person who makes a graph and doesn't copy from another question

allochka39001 [22]

Answer:

H=9;8)

B=(5;4) (ball)

R=(7;0) (hit point)

B'=symetric of B axis perpendicular of x in R

B'=(7+(7-5);4)=(9;4)

Equation BR: y-4=(0-4)/(7-5)(x-5)==>y=-2x+14

Equation RB': y-4=(4-0)/(9-7)(x-9)==>y=2x-14

Is H a point of RB'? y=2x-14 : 8=? 2*9-14==>8=?4 No!

you will not make your putt

Step-by-step explanation:

7 0

2 years ago

The variable Z is directly proportional to X. When X is 5, Z has the value 55.

LuckyWell [14K]

Z = k * x
55 = k * 5
divide both sides by 5
11 = x

z = 11x

z = 11 * 12
z = 132

6 0

2 years ago

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.

7 0

2 years ago

Apple orchard sells Apple's in bags of 10. The orchard sold a total of 2,430 apples one day. How many bags of Apple's was this

zlopas [31]

divide them:

2430/10 = 243 bags

5 0

2 years ago