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3 years ago

How many real number solutions does the equation have? 0=3x2+18x+27

Mathematics

2 answers:

swat323 years ago

7 0

It has only one solution

Morgarella [4.7K]3 years ago

4 0

3(x²+6x+9)=0
3(X+3)(x+3)=0
x=-3 so one solution

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The lengths of the

Dmitrij [34]

Answer:

$S_{\Delta ABC}$ = 4· $S_{\Delta EDC}$

Step-by-step explanation:

From the midpoint theorem, which states that the line that a line drawn such that it joins the midpoints of two sides of a triangle, is parallel to the third side of the triangle and is equal to half the length of the third side

Therefore, the lengths of the sides of ΔDEF, drawn by joining the midpoints of ΔABC is equal to half the length of and parallel to the corresponding side of ΔABC

We therefore, have that the corresponding sides of ΔABC and ΔDEF have a common ratio and a pair of sides in each triangle form same angles, therefore;

ΔDEF is similar to ΔABC by Side, Side, Side SSS similarity.

The length of the perimeter of ΔABC, $S_{\Delta ABC}$ = 2 × The length of the perimeter of triangle ΔEDC, $S_{\Delta EDC}$

$S_{\Delta ABC}$ = 2 × $S_{\Delta EDC}$

∴ $S_{\Delta ABC}$ ≠ 4 × $S_{\Delta EDC}$

The statement which is incorrect is therefore;

$S_{\Delta ABC}$ = 4 × $S_{\Delta EDC}$ .

4 0

3 years ago

The expression sin(x)(cos(x)cot(x)-sin(x)) simplifies to

AysviL [449]

Sin, cos, and tan are fairly difficult questions to answer because if you know what soh, cah, toa is, then 14 million numbers could not even express the amount of stupidity that you possess by asking such a foolish question, young man.

7 0

3 years ago

Read 2 more answers

A half-century ago, the mean height of women in a particular country in their 20s was 64.7 inches. Assume that the heights of

Ainat [17]

Answer:

99.5% of all samples of 21 of today's women in their 20's have mean heights of at least 65.86 inches.

Step-by-step explanation:

We are given that a half-century ago, the mean height of women in a particular country in their 20's was 64.7 inches. Assume that the heights of today's women in their 20's are approximately normally distributed with a standard deviation of 2.07 inches.

Also, a samples of 21 of today's women in their 20's have been taken.

Let $\bar X$ = sample mean heights

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean height of women = 64.7 inches

$\sigma$ = standard deviation = 2.07 inches

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the sample of 21 of today's women in their 20's have mean heights of at least 65.86 inches is given by = P( $\bar X$ $\geq$ 65.86 inches)

P( $\bar X$ $\geq$ 65.86 inches) = P( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ $\geq$ $\frac{65.86-64.7}{\frac{2.07}{\sqrt{21} } }$ ) = P(Z $\geq$ -2.57) = P(Z $\leq$ 2.57)