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3 years ago

PLEASE ANSWER IMMEDIATELY

Mathematics

2 answers:

schepotkina [342]3 years ago

8 0

There are 2 terms

Binomial

insens350 [35]3 years ago

3 0

Binomial because it is the difference of two terms.

Hope this helps!! Let me know if you have ANY questions

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You find a piece of wood that is 10.23 cm long and your friend finds on ethat is 17.65 cm long how long would they be if you att

sleet_krkn [62]

Answer:
27.88

Explanation:
10.23 + 17.65 = 27.88

Hope this helps!!

5 0

3 years ago

Round the number 95,498 to the ten thousands place value

dezoksy [38]

100,000 (sorry if I'm wrong)

8 0

4 years ago

Which ordered pair is in the solution set of the system of linear inequalities? y >3/2 x – 1 y <3/2 x – 1 (–5, 2) (2, 2) (

Vikentia [17]

Answer:

No solution

Option 4 is correct

Step-by-step explanation:

Given: The system of inequality

$y>\dfrac{3}{2}x-1$

$y$

We need to choose correct option.

Option 1: (-5,2)

Put x=-5 and y=2

Equation 1 Equation 2

$2>\dfrac{3}{2}(-5)-1$ $2$

$2>-8.5$ $2$

TRUE FALSE

Option 2: (2,2)

Put x=2 and y=2

Equation 1 Equation 2

$2>\dfrac{3}{2}(2)-1$ $2$

$2>2$ $2$

FALSE FALSE

Option 3: (5,2)

Put x=-5 and y=2

Equation 1 Equation 2

$2>\dfrac{3}{2}(5)-1$ $2$

$2>6.5$ $2$

FALSE TRUE

Option 1, 2 and 3 are incorrect because both equation must be true for solution.

Hence, Option 4 is correct no solution.

6 0

3 years ago

Read 2 more answers

Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1

sleet_krkn [62]

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}$

From Left side:

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)$

$\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}$

NOTE: sin²θ + cos²θ = 1

$\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}$

$\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}$

$\dfrac{2sec^2\theta}{tan^2\theta-1}$

Left side = Right side <em>so proof is complete</em>

8 0

3 years ago

Read 2 more answers

You are playing Monopoly with a friend. She owns and has developed Park Place and Boardwalk with hotels. If you roll an 8, you w

Mariulka [41]

Okay so this friend has a 8, 9, 10 as her options that she pays at.

8: 2,6 9: 6,3 10: 5,5
6,2 3,6 5,5
4,4
4,4

These are the possibilities of rolling these numbers. = 4 possibilities

4/11 (11 because that is the number of possibilities you get for two dice)

that leaves 7/11 possibilities of rolling and not paying!

36% lands on a spot that she pays at and about 64% possibilities of not paying.

Hope this helped!

:)

6 0

3 years ago