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3 years ago

What are the missing terms???

Mathematics

1 answer:

Ivenika [448]3 years ago

6 0

Answer:

$x^{4} +18x^{2} +81=(x^{2} )^{2}+2(9)^{2} +(3)^{4} \\x^{4} +18x^{2} +81=(x^{2} +9)^{2}$

Step-by-step explanation:

This is just a perfect square

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Someone help me please. I don't just want the answer, I need an explanation on how it's done too.

NARA [144]

Answer:

Below.

Step-by-step explanation:

f) (a + b)^3 - 4(a + b)^2

The (a+ b)^2 can be taken out to give:

= (a + b)^2(a + b - 4)

= (a + b)(a + b)(a + b - 4).

g) 3x(x - y) - 6(-x + y)

= 3x( x - y) + 6(x - y)

= (3x + 6)(x - y)

= 3(x + 2)(x - y).

h) (6a - 5b)(c - d) + (3a + 4b)(d - c)

= (6a - 5b)(c - d) + (-3a - 4b)(c - d)

= -(c - d)(6a - 5b)(3a + 4b).

i) -3d(-9a - 2b) + 2c (9a + 2b)

= 3d(9a + 2b) + 2c (9a + 2b)

= 3d(9a + 2b) + 2c (9a + 2b).

= (3d + 2c)(9a + 2b).

j) a^2b^3(2a + 1) - 6ab^2(-1 - 2a)

= a^2b^3(2a + 1) + 6ab^2(2a + 1)

= (2a + 1)( a^2b^3 + 6ab^2)

The GCF of a^2b^3 and 6ab^2 is ab^2, so we have:

(2a + 1)ab^2(ab + 6)

= ab^2(ab + 6)(2a + 1).

4 0

3 years ago

In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a s

Oksi-84 [34.3K]

Answer:

a) $\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

b) From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

c) $P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

$P(Z\geq2.070)=1-P(Z$

Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

$\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4$

And we select a sample size of 64.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

$\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

Part b

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

Part c

For this case we want this probability:

$P(\bar X \geq 285)$

And we can use the z score defined as:

$z=\frac{\bar x -\mu}{\sigma_{\bar x}}$

And using this we got:

$P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

And using a calculator, excel or the normal standard table we have that:

$P(Z\geq2.070)=1-P(Z$

8 0

3 years ago

Need ASAP 20 points !

AlladinOne [14]

Answer:

X = 116

Step-by-step explanation:

Total heptagon angle = 900

124 + 120 + 122 + 136 + 144 + 138 = 784

900 - 784 = 116

4 0

2 years ago

A coach of a baseball team orders hats for the players on his team. Each hat costs $9.95. The shipping charge for the entire ord

maria [59]

Answer:

just do 125.00 -9.95+5.00 and then youll have your answer

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

You are in a hot air balloon looking down at two ponds. Pond A which is in front of your balloon, is at an angle of depression t

MatroZZZ [7]

a. See attachment for the labelled diagram

b. Using the sine ratio, distance from the hot air to pond A is 5,038.9 m, while distance from the hot air to pond B is 6,287.1 m.

c. Distance between the two ponds is 1,263.5 m.

<h3>What is the Sine and Tangent Ratios?</h3>

Sine ratio, is: sin ∅ = opposite side/hypotenuse length

Tangent ratio is: tan ∅ = opposite side/adjacent length.

a. The diagram with the appropriate labels is shown in the image attached below.

b. Use the sine ratio to find the distance from the hot air to pond A (CA) and to pond B (CB):

CA = hypotenuse

∅ = 10°

Opposite = 875 m

sin 10 = 875/CA

CA = 875/sin 10

CA ≈ 5,038.9 m (distance from the hot air to pond A)

CB = hypotenuse

∅ = 8°

Opposite = 875 m

sin 8 = 875/CB

CB = 875/sin 8

CB ≈ 6,287.1 m (distance from the hot air to pond B)

c. Distance between the two ponds, BA = BD + DA.

Apply the tangent ratio to find BD and DA

tan 8 = 875/BD

BD = 875/tan 8

BD = 6,225.9 m

tan 10 = 875/DA

DA = 875/tan 10

DA = 4,962.4 m

Distance between the two ponds = 6,225.9 - 4,962.4 = 1,263.5 m.

Learn more about the sine and tangent ratios on:

brainly.com/question/24137232

#SPJ1

8 0

2 years ago