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myrzilka [38]
1 year ago
12

How does the area of triangle RST compare to the area of triangle LMN? The area of △RST is 2 square units less than the area of

△LMN. The area of △RST is equal to the area of △LMN. The area of △RST is 2 square units greater than the area of △LMN. The area of △RST is 4 square units greater than the area of △LMN.

Mathematics
2 answers:
sashaice [31]1 year ago
8 0
Inscribe triangle RST in the square with dimensions 4×4, as shown in the figure. 

from the area of this square, 4*4=16, we remove the triangles with dimensions 
3×4, 2×1 and 2×4, whose side lengths are shown in the figure, and we are left with the area of triangle RST.


so Area(RTS)=16- \frac{1}{2}*3*4-  \frac{1}{2}*2*1-  \frac{1}{2}*2*4=16-6-1-4=5 units squared

similarly, 

Area(LMN)=4*3- \frac{1}{2}*1*4- \frac{1}{2}*2*2- \frac{1}{2}*2*3=12-2-2-3=5 units squared

Thus, the areas are equal.

Sveta_85 [38]1 year ago
4 0

The answer would be B

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Sum of positive roots of the equation (x2 - 12x + 35) (x2 + 10x + 24) = 504 is
Digiron [165]

Answer:

(3)11

Step-by-step explanation:

We are given that

(x^2-12x+35)(x^2+10x+24)=504

We have to find the sum of positive roots of the equation.

x^2(x^2+10x+24)-12x(x^2+10x+24)+35(x^2+10x+24)=504

x^4+10x^3+24x^2-12x^3-120x^2-288x+35x^2+350x+840=504

x^4-2x^3-61x^2+62x+840-504=0

x^4-2x^3-61x^2+62x+336=0

Factor of 336

2,3,4,6,8,7,

Let x=2

2^4-2^4-61(2^2)+62(2)+336\neq0

x=2 is not the root of equation

x=-2

(-2)^4-2(-2)^3-61(-2)^2+62(-2)+336=0

Hence x=-2 is the root of equation.

x+2 is a factor of equation.

x=3

3^4-2(3^3)-61(3^2)+62(3)+336=0

Therefore, x=3  is the root of equation.

(x+2)(x-3)(x^2-x-56)=0

(x+2)(x-3)(x^2-8x+7x-56)=0

(x+2)(x-3)(x(x-8)+7(x-8))=0

(x+2)(x-3)(x-8)(x+7)=0

x-8=0\implies x=8

x+7=0\implies x=-7

Positive roots are 3 and 8

Sum of positive roots=3+8=11

Option (3) is true.

4 0
1 year ago
Determine the area (in units2) of the region between the two curves by integrating over the x-axis. y = x2 − 24 and y = 1
astra-53 [7]

Answer:

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

Step-by-step explanation:

This case represents a definite integral, in which lower and upper limits are needed, which corresponds to the points where both intersect each other. That is:

x^{2} - 24 = 1

Given that resulting expression is a second order polynomial of the form x^{2} - a^{2}, there are two real and distinct solutions. Roots of the expression are:

x_{1} = -5 and x_{2} = 5.

Now, it is also required to determine which part of the interval (x_{1}, x_{2}) is equal to a number greater than zero (positive). That is:

x^{2} - 24 > 0

x^{2} > 24

x < -4.899 and x > 4.899.

Therefore, exists two sub-intervals: [-5, -4.899] and \left[4.899,5\right]. Besides, x^{2} - 24 > y = 1 in each sub-interval. The definite integral of the region between the two curves over the x-axis is:

A = \int\limits^{-4.899}_{-5} [{1 - (x^{2}-24)]} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} [{1 - (x^{2}-24)]} \, dx

A = \int\limits^{-4.899}_{-5} {25-x^{2}} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} {25-x^{2}} \, dx

A = 25\cdot x \right \left|\limits_{-5}^{-4.899} -\frac{1}{3}\cdot x^{3}\left|\limits_{-5}^{-4.899} + x\left|\limits_{-4.899}^{4.899} + 25\cdot x \right \left|\limits_{4.899}^{5} -\frac{1}{3}\cdot x^{3}\left|\limits_{4.899}^{5}

A = 2.525 -2.474+9.798 + 2.525 - 2.474

A = 9.9\,units^{2}

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

4 0
1 year ago
What are the real zeros of the function g(x) = x^3 + 2x^2 − x − 2?
PSYCHO15rus [73]
If there are real roots to be found for this polynomial, the Rational Root Theorem and synthetic division are the best way to find them. I teach from a book that uses c and d for the possible roots of the polynomial.  C is our constant, 2, and d is the leading coefficient, 1.  The factors of 2 are +/- 1 and +/-2.  The factors for 1 are +/-1 only.  Meaning, in all, there are 4 possibilities as roots for this polynomial.  But there are only 3 total (because our polynomial is a third degree), so we have to find the first one, at least, from our possibilities above.  Let's try x = -1, factor form (x + 1).  If there is no remainder when we do the synthetic division, then -1 is a root.  Put -1 outside the "box" and the coefficients from the polynomial inside: -1  (1  2  -1  -2).  Bring down the first coefficient of 1 and multiply it by the -1 outside to get -1.  Put that -1 up under the 2 and add to get 1.  Multiply 1 times the -1 to get -1 and put that -1 up under the -1 and add to get -2.  -1 times -2 is 2, and -2 + 2 = 0.  So we have our first root of (x+1).  The numbers we get when we do the addition along the way are the coefficients of our new polynomial, the depressed polynomial (NOT a sad one cuz it hates math, but a new polynomial that is one degree less than that of which we started!).  The new polynomial is x^{2} +x-2=0.  That can also be factored to find the remaining 2 roots.  Use standard factoring to find that the other 2 solutions are (x+2) and (x-1).  Our solutions then are x = -2, -1, 1, choice B from above.
6 0
2 years ago
Read 2 more answers
Guillermo is a professional deep water free diver. His altitude (in meters relative to sea level), xxx seconds after diving, is
fgiga [73]

Answer: 140

Step-by-step explanation:

7 0
1 year ago
Read 2 more answers
g In a survey of 500 residents, 300 were opposed to the use of the photo-cop for issuing traffic tickets.The standard error of t
Evgesh-ka [11]

Answer:

Z-score 1.96

Margin of error d= 0.04323

Step-by-step explanation:

Hello!

The study variable in this case is

X: Number of people that oppose the use of photo-cop for issuing traffic tickets in a sample of 500.

n= 500

The parameter of interest is the proportion of people that opposes it.

The estimated proportion is p'= 300/500= 0.6

To estimate the population proportion per Confidence interval you have to approximate the distribution of the sample proportion p' to normal:

p'≈N(p;p(1-p)1/n)

The mean of this distribution is p and the variance is p*(1-p)*1/n

To construct the Confidence interval, since the value of the population proportion is unknown, an estimated variance is used:

p'(1-p')*1/n ⇒ then the estimated standard error is √(p'(1-p')*1/n)

The formula for the confidence interval is:

p' ± Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }

Where "Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }" represents the margin of error of the interval.

Now for a confidence level of 0.95 the value of Z is Z_{0.975}= 1.965

The estimated standard error is already calculated: \sqrt{\frac{p'(1-p')}{n} } = 0.022

The margin of error (d) is then:

d= Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }= 1.965 * 0.022

d= 0.04323

I hope it helps!

7 0
1 year ago
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