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3 years ago

Use the foil method to find the product below. (x+3) (x^2-6x)

Mathematics

1 answer:

Scorpion4ik [409]3 years ago

5 0

x^3 - 3x^2 - 18x

Using the FOIL method, I arrived at my solution!

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to estimate the height of the jin mao tower in shanghai , a tourist sights the top of the tower form a mirror that is 86.7 meter

Ray Of Light [21]

Answer:

The height of the tower = 420.48 meters

Step-by-step explanation:

For better understanding of the solution, see the figure attached below :

Let the height of the tower be x meters

Now, using the laws of reflection : angle of reflection = angle of incidence

Also, both the tower and the tourist are standing parallel to each other

⇒ ∠A = ∠i ( Alternate interior angles are equal)

Similarly, ∠D = ∠r ( Alternate interior angles)

But, ∠i = ∠r

⇒ ∠A = ∠D

Also, the tourist and the tower is perpendicular to the ground surface.

⇒ m∠B = m∠E = 90°

Now, in ΔABC and ΔDEC

∠A = ∠D (Proved above)

m∠B = m∠E = 90°

So, by AA postulate of similarity of triangles, ΔABC ~ ΔDEC

As the sides of similar triangles are proportional to each other

$\frac{AB}{DE}=\frac{BC}{EC}\\\\\implies\frac{1.92}{x}=\frac{0.4}{87.6}\\\\\implies x=\frac{87.6\times 1.92}{0.4}\\\\\bf\implies x = 420.48\textbf{ meters}$

Hence, The height of the tower = 420.48 meters

5 0

3 years ago

What is the value of the 5 in the number 14.5?

WINSTONCH [101]

The alue of 5 is the tenths place.

3 0

3 years ago

PLS ANSWER!!!! which of the following equations corresponds to the table below?

Tju [1.3M]

Answer:

Step-by-step explanation:

just replace x into the equation to see if y match with the table.

5 0

3 years ago

in a random sample of 200 people, 154 said that they watched educational television. fine the 90% confidence interval of the tru

melisa1 [442]

Answer:

[0.7210,0.8189] = [72.10%, 81.89%]

Step-by-step explanation:

The sample size is

n = 200

the proportion is

p = 154/200 = 0.77

Since both np ≥ 10 and n(1-p) ≥ 10

We can approximate this discrete binomial distribution with the continuous Normal distribution. As the sample size is large enough, not applying the continuity correction factor makes no significant difference.

The approximation would be to a Normal curve with this parameters:

Mean

p = 0.77

Standard deviation

$\bf s=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.77*0.23}{200}}=0.0298$

The 90% confidence interval for the proportion would be then

$\bf [0.77-z^*0.0298, 0.77+z^*0.0298]$

where $\bf z^*$ is the 10% critical value for the Normal N(0,1) distribution , this is a value such that the area under the N(0,1) curve outside the interval $\bf [-z^*,z^*]$ is 10%=0.1