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3 years ago

Name an angle of B) 50°

Mathematics

2 answers:

anygoal [31]3 years ago

6 0

Answer:

∠BFC = 50

Step-by-step explanation:

∠AFC = 90

∠AFB + ∠BFC = 90

40 + ∠BFC = 90

∠BFC = 90 - 40

∠BFC = 50

Pani-rosa [81]3 years ago

6 0

Answer: B-C

Step-by-step explanation:

I believe it is the angle of B-C becuase B-E is a strieght line which equals 180. Also EFD looks the same as AFB so it is 40. And DFC is 90 becuase of the square. So we take 180-40-90=50. And the only one that is not solved is B-C

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The perimeter of a rectangular field is 52 yards. The length is 4 more yards than the width. Find the length and width of the fi

hichkok12 [17]

:) Brainliest pls?

Answer:

Width = 11 yards

Length = 15 yards

Step-by-step explanation:

P = 2(l + w)

Consider the width to be x.

The length would then be x+4.

We know that P = 52 yd

52 = 2(x + x + 4)

52 = 2(2x + 4)

52 = 4x + 8 (Subtract 8 from both sides)

44 = 4x (Divide both sides by 4)

x = 11

The width is x, so the width is 11 yards.

The length is x+4, so the length is 15 yards.

8 0

3 years ago

Find the value of the test statistic z using . The claim is that the proportion of adults who smoked a cigarette in the past wee

kolezko [41]

Correct question is;

The claim is that the proportion of adults who smoked a cigarette in the past week is less than 0.35, and the sample statistics include n = 1168 subjects with 385 saying that they smoked a cigarette in the past week. Find the value of the test statistic

Answer:

Test statistic is z = -1.46

Step-by-step explanation:

Let's first of all define the hypotheses:

Null hypothesis:

H0: p = 0.35, i.e 35% in the sample of 1,168 adults have smoked cigarettes in the previous week.

Alternative hypothesis:

Ha: p < 0.35, i.e less than 35% in the sample of 1,168 adults have smoked cigarette in the previous week.

The sample size is, n = 1,168 while the number of adults who smoked in the previous week would be; x = 385

Therefore, the sample proportion of adults who smoked in the previous week would be calculated as;

p^ = x/n = 385/1168 ≈ 0.3296

Now, from Central Limit Theorem for large samples, The sampling distribution of the sample proportion p^, will have a mean of μ = p = 0.35

Formula for standard deviation is;

σ = √[p (1 – p)/n]

σ = √(0.35 × (1 – 0.35)/1168)

σ = √0.0001947774

σ = 0.014

Formula for test statistic is;

z = (p^ - p)/σ

z = (0.3296 - 0.35)/0.014

z = - 1.46

3 0

3 years ago

Ayuda necesito resolver este problema con procedimiento ;)

Paraphin [41]

$x^3-2x^2+x-1$ is one of the prime factors of the polynomial

<h3>How to factor the expression?</h3>

The question implies that we determine one of the prime factors of the polynomial.

The polynomial is given as:

$x^8 - 3x^6 + x^4 - 2x^3 - 1$

Expand the polynomial by adding 0's in the form of +a - a

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^8 -2x^7 + 2x^7 - 4x^6 +x^6 + 2x^5 -2x^5- 3x^4 + 4x^4 + 2x^3 -6x^3+2x^3- x^2 -3x^2 +4x^2-2x+2x-1$

Rearrange the terms

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^8 -2x^7 + 2x^5 - 3x^4 + 2x^3 - x^2 + 2x^7 - 4x^6 + 4x^4 -6x^3+4x^2-2x+x^6-2x^5+2x^3-3x^2+2x-1$

Factorize the expression

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^2(x^6-2x^5+2x^3-3x^2+2x-1) + 2x(x^6-2x^5+2x^3-3x^2+2x-1) + 1(x^6-2x^5+2x^3-3x^2+2x-1)$

Factor out x^6-2x^5+2x^3-3x^2+2x-1

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x^2+2x + 1)(x^6-2x^5+2x^3-3x^2+2x-1)$

Express x^2 + 2x + 1 as a perfect square

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6-2x^5+2x^3-3x^2+2x-1)$

Expand the polynomial by adding 0's in the form of +a - a

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6- 2x^5+x^4-x^4-x^3 +x^3-2x^3-x^2 -2x^2 +x+x - 1)$

Rearrange the terms

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6- 2x^5+x^4-x^3-x^4-2x^3-x^2+x+x^3-2x^2 +x - 1)$

Factorize the expression

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^3(x^3-2x^2+x-1) -x(x^3-2x^2+x-1)+1(x^3-2x^2+x-1))$

Factor out x^3-2x^2+x-1

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^3 -x+1)(x^3-2x^2+x-1)$

One of the factors of the above polynomial is $x^3-2x^2+x-1$ .

This is the same as the option (c)

Hence, $x^3-2x^2+x-1$ is one of the prime factors of the polynomial