MATHEMATICS COLLEGE

Determine the shaded areas part 2

Answers

Answer 1

Answer:

9514 1404 393

Answer:

(a) cannot be determined

(b) 44 cm^2

(d) 180 m^2

(e) 132 m^2

Step-by-step explanation:

(a) missing a horizontal dimension

(b) The difference between the bounding rectangle and the lower-left cutout is ...

(8 cm)(7 cm) -(3 cm)(4 cm) = (56 -12) cm^2 = 44 cm^2

(13 m)(7 m) -(4 m)(1 m) = (91 -4) m^2 = 87 m^2

(d) The difference between the bounding rectangle and the two cutouts is ...

(20 m)(25 m) -(16 m)(20 m) = (20 m)(25 -16) m = (20 m)(9 m) = 180 m^2

(e) The difference between the bounding rectangle and the two cutouts is ...

(14 m)(12 m) -(12 m)(3 m) = (12 m)(14 -3) m = (12 m)(11 m) = 132 m^2

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2x- y = -2
2x + 4y =8

Answers

Answer:

x=8

y=4

Step-by-step explanation:

Answer:

Your point would be: (0,2)

x = 0

y = 2

Step-by-step explanation:

This would be the case if you're looking for the point by solving with substitution, or any other methods, of the systems of equations.

In any case, you are looking to find either x or y first.

I will demonstrate with the 'y' value, since it is easier when cancelling out the 'x' value.

2x - y = -2

2x + 4y = 8

(I want to make one of the two equations negative, so that I can eliminate or cancel out the 'x' value(s)).

- (2x - y = -2) --> -2x + y = 2

2x +4y = 8

---------------------------------------

5y = 10

--- ---

5 5

------------

y = 2

(Reason for putting a negative in front of the equation '2x - y = -2' was made in order to eliminate the value(s) '2x' in both equation, and in order to do that, one equation must be negative, while the other is positive, for example:

x + 2y = 6

x + 3y = 5

(You can make any of these equations negative, but only one equation is needed to be negative for the value you want to cancel out,

DO NOT TRY AND CANCEL OUT BOTH X AND Y)

-(x + 2y = 6)

Turn into:

-x - 2y = -6

x + 3y = 5

---------------

y = -1

(Back to the Original Problem)

We now need to find the 'x' value(s):

4(2x - y = -2)

8x - 4y = -8

2x + 4y = 8

----------------

10x = 0

--- ---

10 10

----------------

x = 0

Your final answer will be:

(0,2)

We have 30 miles of fencing which can be used to enclose a rectangular piece of grazing land along a straight portion of a river. No fence is required along the river. The grazing land will be subdivided into two sections by means of a fence parallel to the sides and perpendicular to the river. Write a function that expresses the total area in terms of the width x of the side of the grazing area perpendicular to the straight river.

Answers

Answer:

The area of the gazing land is $\frac{30x-x^2}2$ square miles.

Step-by-step explanation:

Given that, 30 miles of fencing can be used to enclose a rectangular piece of grazing land along a straight piece of river.

Let the length of the rectangular piece of gazing which is along the river be y and the width of the rectangular piece of gazing be x.

Along the river side, no fence is required.

Therefore total length of fence is = 2(x+y)-x

=2y+x

∴2y+x=30

⇒2y=30-x

$\Rightarrow y=(30-x)/(2)$

The area of the rectangular piece of gazing is = Length×width

=xy

$=x.\frac{30-x}2$

$=\frac{30x-x^2}2$ square miles.

Can you guys help me to find the measure of each angle tysm.

Answers

Answer:

∠EBF = 51°

∠DBE = 17°

∠ABF = 141°

∠EBA = 90°

∠DBC = 107°

∠DBF = 68°

Step-by-step explanation:

Hope this helps

Answer:

a. m<EBF = 90 - 39 = 51

b. m<EBA = 90

c. m<DBE = 90 - 73 = 17

d. m<DBC = 90 + 17 = 107

e. m<ABF = 90 + 51 = 141

f. m<DBF = 17 + 51 = 68

<ABD, <DBE, <EBF, <FBC, <DBF

<ABF, <DBC

<ABE, <EBC

HELP PLEASE ASAP What is the length of the diagonal of the square shown below?

Answers

Answer:

C) diagonal = 5√2

Step-by-step explanation:

diagonal² = 5² + 5²

diagonal² = 50

diagonal = √50

diagonal = √25x2

diagonal = 5√2

Answer:

C) diagonal = 5√2

Step-by-step explanation:

diagonal² = 5² + 5²

diagonal² = 50

diagonal = √50

diagonal = √25x2

diagonal = 5√2

The hypotenuse of a right triangle is 10 meters and its legs measure x and x + 2. How long are the legs?

Answers

Solution :

Let's solve by using Pythagoras theorem,

$\hookrightarrow \: (10) {}^(2) = {(x})^(2) + (x + 2) {}^(2)$

$\hookrightarrow \: 100 = {x}^(2) + {x}^(2) + 4x + 4$

$\hookrightarrow \: 100 - 4 = 2 {x}^(2) + 4x$

$\hookrightarrow \: 96 = 2 {x}^(2) + 4x$

$\hookrightarrow \: 2 {x}^(2) + 4x - 96 = 0$

$\hookrightarrow \: 2( {x}^(2) + 2x - 48) = 0$

$\hookrightarrow \: {x}^(2) + 2x - 48 = 0$

$\hookrightarrow \: {x}^(2) + 8x - 6x - 48 = 0$

$\hookrightarrow \: x(x + 8) - 6(x + 8) = 0$

$\hookrightarrow \: (x + 8)(x - 6) = 0$

now there are two cases

Case 1 : when (x + 8) = 0

$\hookrightarrow \: x + 8 = 0$

$\hookrightarrow \: x = - 8$

but the value of x can't be negative, since side of a triangle isn't a negative value .

Case 2 : when (x - 6) = 0

$\hookrightarrow \: x - 6 = 0$

$\hookrightarrow \: x = 6$

therefore the measure of other two sides are :

$\hookrightarrow \: x = 6 \: units$

$\hookrightarrow \: x + 2 = 6 + 2 = 8 \: units$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathrm{TeeNForeveR }$

NBC News reported on May 2, 2013, that 1 in 20 children in the United States have a food allergy of some sort. Consider selecting a random sample of 15 children and let X be the number in the sample who have a food allergy. Then X ~ Bin(15, 0.05). (Round your probabilities to three decimal places.) (a) Determine both P(X ≤ 3) and P(X < 3). P(X ≤ 3) = P(X < 3) = (b) Determine P(X ≥ 4). P(X ≥ 4) = (c) Determine P(1 ≤ X ≤ 3). P(1 ≤ X ≤ 3) = (d) What are E(X) and σX? (Round your answers to two decimal places.) E(X) = σX = (e) In a sample of 90 children, what is the probability that none has a food allergy?

Answers

Answer:

a) P(X ≤ 3) = 0.9946

P(X < 3) = 0.9639

b) P(X ≥ 4) = 0.0054

c) P(1 ≤ X ≤ 3) = 0.5313

d) E(X) = 0.75

σX = 0.84

e) P(X=0) = 0.0099

Step-by-step explanation:

We have x: number in the sample who have a food allergy. As the sample is of n=15 and p=0.05, we have:

$X \sim Bin(15, 0.05)$

a) We have to determine P(X ≤ 3) and P(X < 3)

We can calculate P(X ≤ 3) as the sum of P(0), P(1), P(2) and P(3).

$P(x\leq 3)=\sum_(k=0)^3P(k)\n\n\nP(x=0) = \binom{15}{0} p^(0)q^(15)=1*1*0.4633=0.4633\n\nP(x=1) = \binom{15}{1} p^(1)q^(14)=15*0.05*0.4877=0.3658\n\nP(x=2) = \binom{15}{2} p^(2)q^(13)=105*0.0025*0.5133=0.1348\n\nP(x=3) = \binom{15}{3} p^(3)q^(12)=455*0.0001*0.5404=0.0307\n\n\nP(x\leq 3)=0.4633+0.3658+0.1348+0.0307=0.9946$