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3 years ago

D^2 + 2d - 8 = 0 explain plz

Mathematics

1 answer:

schepotkina [342]3 years ago

7 0

Answer:

d^2 +2d-8=0

first pick two numbers which add up to make +2 and by multiplying those two numbers they should be able to create -8.

so for example 4 and 2

4-2=2

4 x -2= -8

so now factorise:

d^2 +2d-8=0

d^2+4d-2d -8=0

d(d+4)-2(d+4)=0

d+4=0 d-2=0

d=-4 d=2

Step-by-step explanation:

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Write an equation of an ellipse in standard form with the center at the origin and a height of 12 units and width of 19 units.

gladu [14]

Answer:

The equation of the ellipse in standard form is 4x²/361 + y²/36 = 1

Step-by-step explanation:

* Lets revise the equation of the ellipse

- The standard form of the equation of an ellipse with center (0 , 0 )

and major axis parallel to the x-axis is x²/a² + y²/b² = 1

# a > b

- The length of the major axis is 2a

- The coordinates of the vertices are ( ± a , 0 )

- The length of the minor axis is 2b

- The coordinates of the co-vertices are ( 0 , ± b )

- The coordinates of the foci are ( ± c , 0 ) , where c ² = a ² − b²

* Lets solve the problem

∵ The center of the ellipse is (0 ,0)

∵ Its width is 19 units

∴ The length of the major axis is = 19

∴ 2a = 19 ⇒ divide both sides by 2

∴ a = 19/2 ⇒ ∴ a² = 361/4

∵ Its height is 12 units

∴ The length of the minor axis is = 12

∴ 2b = 12 ⇒ divide both sides by 2

∴ b = 12/2 = 6 ⇒ ∴ b² = 36

- Lets write the equation in standard form

∵ The equation is x²/a² + y²/b² = 1

∴ x²/(361/4) + y²/36 = 1 ⇒ simplify it

∴ 4x²/361 + y²/36 = 1

* The equation of the ellipse in standard form is 4x²/361 + y²/36 = 1

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3 years ago

Solve by Substitution. 3x+2y=-4 y=4x-2

qwelly [4]

x=0

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You'd put them in an ordered point, so (0,-2).

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3 years ago

How to know if an equation is false, true, or an open scentence

beks73 [17]

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A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. (a) What must be true

valentina_108 [34]

Answer:

(a) The distribution of the sample mean ( $\bar x$ ) is N (68, 4.74²).

(b) The value of $P(\bar X$ is 0.7642.

Step-by-step explanation:

A random sample of size n = 10 is selected from a population.

Let the population be made up of the random variable X.

The mean and standard deviation of X are:

$\mu=68\\\sigma=15$

(a)

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. n = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

$\mu_{\bar x}=\mu=68$