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3 years ago

HELPPPPPO

Mathematics

1 answer:

Rina8888 [55]3 years ago

6 0

Answer:

150

Step-by-step explanation:

'x' = number of students out of 500 who selected math

18/60 = x/500

cross-multiply:

60x = 9000

x = 150

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Find an nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing utili

Marrrta [24]

We need to find a polynomial function f(x) with degree 3 and zeros 2 and 2i, such that:

$f(-1)=30$

Since 2i is a zero, -2i is also a zero of this function. Thus, we have:

$f(x)=a(x-2)(x-2\imaginaryI)(x+2\imaginaryI)$

where a is a constant.

Expanding the expression on the right side, we obtain:

$\begin{gathered} \\ f(x)=a(x-2)\lbrack x²-(2i)²\rbrack \\ \\ f(x)=a(x-2)(x²-4i²),\text{ i^^b2=-1} \\ \\ f(x)=a(x-2)(x²+4) \\ \\ f(x)=a(x³-2x²+4x-8) \end{gathered}$

Now, using f(-1) = 30, we obtain:

$\begin{gathered} 30=a\lbrack(-1)³-2(-1)²+4(-1)-8\rbrack \\ \\ 30=a(-1-2-4-8) \\ \\ 30=a(-15) \\ \\ a=\frac{30}{-15} \\ \\ a=-2 \end{gathered}$

Therefore, the function is:

Answer

$f(x)=-2x^{3}+4x^{2}-8x+16$

6 0

1 year ago

Emelina wrote the equation of a line in point-slope form as shown below.

Harman [31]

Answer: $y=3x+2$

Step-by-step explanation:

Given: Emelina wrote the equation of a line in point-slope form as shown below.

$(y+4)=3(x+2)$

To write the equation in intercept form, first we multiply 3 inside the bracket values in the right side , we get

$y+4=3x+6$

Now, subtract 4 from both sides , we get

$y=3x+2$

Hence, Emelina’s equation in slope-intercept form : $y=3x+2$

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4 years ago

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In the number 5,789 which digit is in the hundreds place?

bogdanovich [222]

Answer:

Step-by-step explanation:

6 0

4 years ago

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write the equation of the line that passes through the point (-1,-6) and is perpendicular to a line that passes through the poin

Burka [1]

Answer:

$y=\frac{2}{3}x-\frac{16}{3}$

Step-by-step explanation:

step 1

Find the slope of the given line

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have

(-2,5) and (-4,8)

substitute the values in the formula

$m=\frac{8-5}{-4+2}$

$m=\frac{3}{-2}$

$m=-\frac{3}{2}$

step 2

Find the slope of the perpendicular line to the given line

we know that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

$m_1*m_2=-1$

$m_1=-\frac{3}{2}$ ----> slope of the given line

therefore

$m_2=\frac{2}{3}$ ---> slope of the perpendicular line to the given line

step 2

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=\frac{2}{3}$

$point\ (-1,-6)$

substitute

$y+6=\frac{2}{3}(x+1)$

step 3

Convert to slope intercept form

$y=mx+b$

isolate the variable y

$y+6=\frac{2}{3}x+\frac{2}{3}$

$y=\frac{2}{3}x+\frac{2}{3}-6$

$y=\frac{2}{3}x-\frac{16}{3}$

4 0

4 years ago

Helpppppppppppppp

Genrish500 [490]

Answer: Vertical angles are congruent.

Step-by-step explanation:

In the given picture, NS and RT are intersecting line segments, intersects at point Q.

We know that when two lines intersect to make an cross, angles on opposite sides of the cross are called vertical angles. These angles are equal by the theorem that says Vertical angles are congruent.

⇒∠NQT ≅ ∠SQR

Therefore, the reason that justifies the step 2 :- ∠NQT ≅ ∠SQR is "Vertical angles are congruent" in the proof.

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4 years ago

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