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3 years ago

If m□ebd=4x-8 and m□ebc=5x+20,find the value of x and m□ebc

Mathematics

1 answer:

RoseWind [281]3 years ago

5 0

m□ebd=4 x-8 and m□ebc=5 x+20

This is solvable only if e b is the initial side and b d and b c lies on opposite side of each other and lies on a line i.e c,b,d are Collinear.

∠ebd and ∠ebc will form a linear pair.The meaning of linear pair is that angles forming on one side of a straight line through a common vertex which are adjacent is 180°.

i.e

∠ ebd + ∠ebc = 180°

4 x- 8 + 5x + 20= 180°

adding like terms

⇒ 9 x +12 =180°

⇒ 9 x = 180° - 12

⇒ 9 x = 168°

⇒ x =( 168/9)°=(56/3)°

now m□ebc =5 x +20

= 5 × 56/3 + 20

= 280/3 + 20

=340/3

m□ebc=( 340/3)°

So, solution set is x =(56/3)° and m□ebc =(340/3)°

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Which equation involves a prime quadratic and cannot be solved by factoring? A. x2 + 6x + 9 = 0 B. x2 − x − 6 = 0 C. x2 + 3x − 4

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3 years ago

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All are correct except these 2:

The graph has one x-intercept:
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No, the y-intercept is at (0,5)

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3 years ago

Nitrates are groundwater contaminants derived from fertilizer, septic tank seepage and other sewage. Nitrate poisoning is partic

o-na [289]

Answer:

a) $n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.96})^2}=2401$

And rounded up we have that n=2401

b) $n=\frac{0.07(1-0.07)}{(\frac{0.02}{1.96})^2}=625.22$

And rounded up we have that n=626

And we see that if we have previous info about p the sample size varies a lot.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

Part a

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.02$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

Since we don't have prior information about the population proportion we can use ase estimator $\hat p =0.5$ . And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.96})^2}=2401$

And rounded up we have that n=2401

Part b

For thi case the estimator for p is $\hat p =0.07$

$n=\frac{0.07(1-0.07)}{(\frac{0.02}{1.96})^2}=625.22$

And rounded up we have that n=626

And we see that if we have previous info about p the sample size varies a lot.

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3 years ago

The game of Connex contains one 4-unit piece, two identical 3-unit pieces, three identical 2-unit pieces and four identical 1-un

Aleks [24]

Answer: 277 ways

Step-by-step explanation:

Let’s start bycreating 10-unit pieces using the 4-unit piece.

The arrangements are:

1). 4-3-3 (3 permutations)

2). 4-3-2-1 = 4! = 24 permutations.

3). 4-3-1-1-1 (5*[4!/(3!1!)]

= 5*4

= 20permutations

4). 4-2-2-2 (4 permutations)

5). 4-2-2-1-1 (5 *[4!/(2!2!)]

= 5*6

= 30 permutations

6). 4-2-1-1-1-1(6*[5!/(4!1!)]

= 6*5

= 30 permutations

Let’s-consider the arrangements using one or more3-unit pieces and no 4-unit piece:

7). 3-3-2-2 (4!/(2!2!)

8). 3-3-2-1-1 (5*(4!/(2!2!)

= 5*6

= 30 permutations.

9). 3-3-1-1-1-1 (6!/(4!2!) = 0

10). 3-2-2-2-1 (5*4!/(3!1!)

= 5*4

= 20permutations

11). 3-2-2-1-1-1 (6*5!/(3!2!)

= 6*10

= 60 permutations

Finally we would look at the arrangements using only 1-and 2-unit pieces:

12). 2-2-2-1-1-1-1 (7!/(4!3!)

= 35 permutations

add them all up:

(3 + 24 + 20 + 4) + (30 + 30 + 6 + 30) + (15 + 20 + 60 + 35)

=51 + 96 + 130

= 277ways

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2 years ago

How do you answer this?

SVEN [57.7K]

$(m^{-2})^5=m^{-2\cdot5}=m^{-10}$

$x^2+3x-10=\\
x^2+5x-2x-10=\\
x(x+5)-2(x+5)=\\
(x-2)(x+5)$

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3 years ago

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