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Yakvenalex [24]
3 years ago
13

Please help me solve this.

Mathematics
1 answer:
ololo11 [35]3 years ago
8 0

Answer:

(x+1)(x+7)

Step-by-step explanation:

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Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the zeros −3 and 5
love history [14]

Answer:

<em>f(x) = x² (2+i)x-15-3i</em>

Step-by-step explanation:

Since the zeros of the equation are -3 and 5+i, hence the factors of the polynomial in x is (x+3) and (x-(5+i))

Multiplying both factors

f(x) =  (x+3)(x-(5+i))

f(x) = (x²-(5+i)x+3x -3(5+i))

f(x) = x² - (5+i- 3)x -15-3i

f(x) = x² (2+i)x-15-3i

<em>hence the required polynomial function in x is f(x) = x² (2+i)x-15-3i</em>

7 0
3 years ago
Andrew uses 630 one-inch unit cubes to completely fill the inside of a rectangular box. Which could be the dimensions of the box
dsp73

we have 630 one-inch unit cubes and we want to completely fill the rectangular box (unknown dimensions).

If all the cubes are fitted tightly inside rectangular box without living any space, then box volume would be equal to cubes volume.

There are 630 one-inch unit cubes, so volume of cubes = 630 cubic inches.

Now the volume of rectangular box would also be 630 cubic inches.

We know the formula for volume of rectangular box = length × width × height.

So we need to find any three positive integers whose product is 630.

Out of all given choices, only option A satisfies the condition of factors of 630.

Hence, option A i.e. (7 in x 9 in x 10 in) is the final answer.

7 0
3 years ago
I'm not sure how you spell it, but it's the abbreviation for the Socials, the jet set, the West-side rich kids. It's like the te
ale4655 [162]

Answer:

I think it is B

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
Solve 2cos ²y -siny -1=0 for 0° ≤y≤360°
natima [27]

 

\displaystyle\bf\\2cos^2y -sin\,y -1=0~~~for~~0^o\leq y \leq360^o\\\\cos^2y=1-sin^2y\\\\2(1-sin^2y) -sin\,y -1=0\\\\2-2sin^2y-sin\,y-1=0\\\\-2sin^2y-sin\,y+2-1=0\\\\-2sin^2y-sin\,y+1=0~~~\Big|\times(-1)\\\\2sin^2y+sin\,y-1=0

.

\displaystyle\bf\\\boxed{\bf sin\,y=x}\\\\2x^2+x-1=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-1\pm\sqrt{1-4\cdot2\cdot(-1)}}{2\cdot2}=\\\\=\frac{-1\pm\sqrt{1+8}}{4}=\frac{-1\pm\sqrt{9}}{4}=\frac{-1\pm3}{4}\\\\x_1=\frac{-1+3}{4}=\frac{2}{4}=\boxed{\bf\frac{1}{2}}\\\\x_2=\frac{-1-3}{4}=\frac{-4}{4}=\boxed{\bf-1}\\\\sin\,y=\frac{1}{2}\\\\\boxed{\bf y_1=\frac{\pi}{6}~~or~~(30^o)}\\\\\boxed{\bf y_2=\frac{5\pi}{6}~~or~~(150^o)}\\\\sin\,y=-1\\\\\boxed{\bf y_3=\frac{3\pi}{2}~~or~~(270^o)}

 

 

5 0
2 years ago
What is the area of the shaded portion? 18 points to whomever answers this.
iVinArrow [24]
The second answer
 pretty sure
8 0
2 years ago
Read 2 more answers
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