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3 years ago

What's 18y^3-6y^2-63y+21

Mathematics

2 answers:

xeze [42]3 years ago

5 0

18y^(3)-6y^(2)-63y+21 solve by factoring

amid [387]3 years ago

5 0

Let's simplify step-by-step.
18y3−6y2−−63y+21
=18y3+−6y2+63y+21
Answer:
=18y3−6y2+63y+21

Hope This Helps!

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Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar. Aakash has a fever.

Anna11 [10]

Answer:

Step-by-step explanation:

Aakash body temperature during the day = 99.2°F.

If by evening his body temperature increases by 3.4°F, then his body temperature in the evening will be expresses as the sum of his temperature in the day and the increment.

Temperature in the evening = 99.2°F+3.4°F.

Temperature in the evening = 102.6°F.

Hence his body temperature in the evening is 102.6°F.

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3 years ago

Costs of homes can be very different in different parts of the United States. a. A 450-square-foot apartment in New York City co

VLD [36.1K]

Answer:

The cost of homes per square foot differs in different part of the United states. The cost of homes per square foot is gotten by using the formula:

cost of homes per square foot = Cost of apartment / area of apartment

1) Cost of apartment = $540000, area of apartment = 450 ft²

cost of homes per square foot = Cost of apartment / area of apartment = $540000/ 450 ft² = $1200 / ft²

2) cost of homes per square foot = $110/ ft², area of apartment = 2100 ft²

cost of homes per square foot = Cost of apartment / area of apartment

$110 / ft² = Cost of apartment / 2100 ft²

Cost of apartment = $110 / ft² × 2100 ft² = $231000

Step-by-step explanation:

Hope this helps plz mark brainiest

5 0

3 years ago

Read 2 more answers

Given points A(-1, -2) and B(2, 4) where AP: BP=1:2, find the locus of point P.

koban [17]

Answer:

$x^2 + 4x + y^2 +8y = 0$

Step-by-step explanation:

Given

$A = (-1,-2)$

$B = (2,4)$

$AP:BP = 1 : 2$

Required

The locus of P

$AP:BP = 1 : 2$

Express as fraction

$\frac{AP}{BP} = \frac{1}{2}$

Cross multiply

$2AP = BP$

Calculate AP and BP using the following distance formula:

$d = \sqrt{(x - x_1)^2 + (y - y_1)^2}$

So, we have:

$2 * \sqrt{(x - -1)^2 + (y - -2)^2} = \sqrt{(x - 2)^2 + (y - 4)^2}$

$2 * \sqrt{(x +1)^2 + (y +2)^2} = \sqrt{(x - 2)^2 + (y - 4)^2}$

Take square of both sides

$4 * [(x +1)^2 + (y +2)^2] = (x - 2)^2 + (y - 4)^2$

Evaluate all squares

$4 * [x^2 + 2x + 1 + y^2 +4y + 4] = x^2 - 4x + 4 + y^2 - 8y + 16$

Collect and evaluate like terms

$4 * [x^2 + 2x + y^2 +4y + 5] = x^2 - 4x + y^2 - 8y + 20$

Open brackets

$4x^2 + 8x + 4y^2 +16y + 20 = x^2 - 4x + y^2 - 8y + 20$

Collect like terms

$4x^2 - x^2 + 8x + 4x + 4y^2 -y^2 +16y + 8y + 20 - 20 = 0$

$3x^2 + 12x + 3y^2 +24y = 0$

Divide through by 3

$x^2 + 4x + y^2 +8y = 0$

3 0

3 years ago

Find three numbers such that their sum is 12, the sum of the first, twice the second, and three times the third is 31, and the s

Tpy6a [65]

Answer:

a = 3, b = -1, c = 10

Step-by-step explanation:

Let the three numbers be a, b and c.

Equation 1: a + b + c = 12

Equation 2: a + 2b + 3c = 31

Equation 3: 9b + c = 1

Equation 2 - Equation 1:

Equation 4: b + 2c = 19

Equation 3 times by the number 2

Equation 5: 18b + 2c = 2

Equation 5 - Equation 4

17b = -17

b = -1

Substitute into Equation 4:

2c - 1 = 19

2c = 20

c = 10

Substitute into Equation 1:

a + b + c = 12

a - 1 + 10 = 12

a = 3

8 0

3 years ago

Read 2 more answers

Given sin(−θ)=−1/6 and tanθ=−√35/35 What is the value of cosθ?

murzikaleks [220]

Answer:

$cos(\theta)=-\frac{\sqrt{35} }{6}$

Step-by-step explanation:

Recall the negative angle identity for the sine function:

$sin(- \theta)=-sin(\theta)$

Then, we can find the value of $sin(\theta)$ :

$sin(\theta)=-sin(-\theta)\\sin(\theta) =-(-\frac{1}{6} )\\sin(\theta)= \frac{1}{6}$

Now recall the definition of the tangent function:

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

Therefore, now that we know the value of $sin(\theta)$ , we can solve in this equation for $cos(\theta)$

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}\\-\frac{\sqrt{35} }{35} =\frac{1/6}{cos(\theta)} \\cos(\theta)=-\frac{\frac{1}{6} }{\frac{\sqrt{35} }{35} } \\cos(\theta)=-\frac{35}{6\,\sqrt{35} } \\cos(\theta)=-\frac{\sqrt{35} }{6}$

5 0

3 years ago