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EleoNora [17]
3 years ago
10

A triangle has an area of 9m2 and a base of 6m. A. Determine the value of the height of this triangle. Explain your steps.

Mathematics
1 answer:
Genrish500 [490]3 years ago
7 0
Area = 1/2 x b x h

9 = 1/2 x 6 x h

18 = 6 x h 
h = 18/6
h = 3
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c. 50

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Read 2 more answers
When an electric current passes through two resistors with resistance r1 and r2, connected in parallel, the combined resistance,
kondaur [170]

Answer:

a)

The combined resistance of a circuit consisting of two resistors in parallel is given by:

\frac{1}{R}=\frac{1}{r_1}+\frac{1}{r_2}

where

R is the combined resistance

r_1, r_2 are the two resistors

We can re-write the expression as follows:

\frac{1}{R}=\frac{r_1+r_2}{r_1r_2}

Or

R=\frac{r_1 r_2}{r_1+r_2}

In order to see if the function is increasing in r1, we calculate the derivative with respect to r1: if the derivative if > 0, then the function is increasing.

The derivative of R with respect to r1 is:

\frac{dR}{dr_1}=\frac{r_2(r_1+r_2)-1(r_1r_2)}{(r_1+r_2)^2}=\frac{r_2^2}{(r_1+r_2)^2}

We notice that the derivative is a fraction of two squared terms: therefore, both factors are positive, so the derivative is always positive, and this means that R is an increasing function of r1.

b)

To solve this part, we use again the expression for R written in part a:

R=\frac{r_1 r_2}{r_1+r_2}

We start by noticing that there is a limit on the allowed values for r1: in fact, r1 must be strictly positive,

r_1>0

So the interval of allowed values for r1 is

0

From part a), we also said that the function is increasing versus r1 over the whole domain. This means that if we consider a certain interval

a ≤ r1 ≤ b

The maximum of the function (R) will occur at the maximum value of r1 in this interval: so, at

r_1=b

6 0
3 years ago
Given: KLMN is a trapezoid m∠N = m∠KML ME ⊥ KN , ME = 3 5 , KE = 8, LM KN = 3 5 Find: KM, LM, KN, Area of KLMN
prisoha [69]
A) Find KM∠KEM is a right angle hence ΔKEM is a right angled triangle Using Pythogoras' theorem where the square of hypotenuse is equal to the sum of the squares of the adjacent sides we can answer the 
KM² = KE² + ME²KM² = 8² + (3√5)²       = 64 + 9x5KM = √109KM = 10.44
b)Find LMThe ratio of LM:KN is 3:5 hence if we take the length of one unit as xlength of LM is 3xand the length of KN is 5x ∠K and ∠N are equal making it a isosceles trapezoid. A line from L that cuts KN perpendicularly at D makes KE = DN
KN = LM + 2x 2x = KE + DN2x = 8+8x = 8LM = 3x = 3*8 = 24
c)Find KN Since ∠K and ∠N are equal, when we take the 2 triangles KEM and LDN, they both have the same height ME = LD.
∠K = ∠N  Hence KE = DN the distance ED = LMhence KN = KE + ED + DN since ED = LM = 24and KE + DN = 16KN = 16 + 24 = 40
d)Find area KLMNArea of trapezium can be calculated using the  formula below Area = 1/2 x perpendicular height between parallel lines x (sum of the parallel sides)substituting values into the general equationArea = 1/2 * ME * (KN+ LM)          = 1/2 * 3√5 * (40 + 24)         = 1/2 * 3√5 * 64         = 3 x 2.23 * 32         = 214.66 units²
7 0
3 years ago
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