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3 years ago

Determine the exact value

Mathematics

1 answer:

adell [148]3 years ago

6 0

Answer:(theta) =π only

Step-by-step explanation:

-4cos(theta)+1=5

Subtract 1 from both sides

-4cos(theta)+1-1=5-1

-4cos(theta)=4

Divide both sides by -4

-4cos(theta) ➗ -4=4 ➗ -4

cos(theta)=-1

(theta)=cos(inverse)(-1)

(theta)=180

(theta)=π

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What is the factored form of x^12y^18+1

Black_prince [1.1K]

Answer: The required factored form of the given expression is

$(x^4y^6+1)(x^8y^{12}-x^4y^6+1).$

Step-by-step explanation: We are given to find the factored form of the following algebraic expression:

$E=x^{12}y^{18}+1.$

We will be using the following formula:

$a^3+b^3=(a+b)(a^2+ab+b^2).$

Now, we have

$E\\\\=x^{12}y^{18}+1\\\\=(x^4y^6)^3+1^3\\\\=(x^4y^6+1)\{(x^4y^6)^2-x^4y^6\times1+1^2\}\\\\=(x^4y^6+1)(x^8y^{12}-x^4y^6+1).$

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3 years ago

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What is the distance between the points P(-5, 4), Q(7, -5)

Nikolay [14]

Answer:

Step-by-step explanation:

Distance between two points is = √(x2 - x1)² + (y2 - y1)²

x1 = -5, y1 = 4; x2 = 7, y2 = -5

|PQ| = √(7 - (-5))² + (-5 - 4)²

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3 years ago

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I think this is the answer

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3 years ago

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Ms. Lady works-out with Ms. Beetle; her average crawling time has increased to 4 feet per minute, while Ms. Beetle, who is used

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Answer:

Senn needs to give Ms. Lady 4.6 minutes head start

Senn needs to give Ms. Beetle 18.4 minutes head start

The equation for each bug are;

For Senn, the equation for the experiment is 5 × t₁ = 92 feet

For Ms. Lady, the equation for the experiment is 4 × (t₁ + 4.6) = 92 feet

For Ms. Beetle , the equation for the experiment is 4 × (t₁ + 18.4) = 92 feet

Please find attached the required graph and table of values

Step-by-step explanation:

The given parameters are;

The average crawling time of Ms. Lady = 4 feet per minute

The average crawling time of Ms. Beetle = 2.5 feet per minute

The average crawling time of Senn = 5 feet per minute

The distance to the rose bush = 92 feet

Therefore, we have;

The time, t, duration for Senn to arrive at the Rose bush is given by the following relation,

Time, t = Distance, d/(Speed, s)

Given that the speed of the bugs is equal to their average crawling time, we have

For Senn

Time, t = 92/(5 ft/Min) = 18.4 minutes

For, Ms. Lady

Time, t = 92/(4 ft/Min) = 23 minutes

For, Ms. Beetle

t = 92/(2.5 ft/Min) = 36.8 minutes

Therefore;

Senn needs to give Ms. Lady 23 - 18.4 = 4.6 minutes head start

Similarly, Senn needs to give Ms. Beetle 36.8 - 18.4 = 18.4 minutes head start

The equation for each bug are therefore;

For Senn, the equation for the experiment is given as follows ;

5 × t₁ = 92 feet

For Ms. Lady, the equation for the experiment is given as follows;

4 × (t₁ + 4.6) = 92 feet

For Ms. Beetle , the equation for the experiment is given as follows;

4 × (t₁ + 18.4) = 92 feet.

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Answer:

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