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Sidana [21]
3 years ago
7

Multiply 5x^2-6x+2 4x^2-3x

Mathematics
2 answers:
mixas84 [53]3 years ago
8 0

Answer:

20x^{4}-39x^{3} +26x^{2}-6x\\

Step-by-step explanation:

Multiply the two polynomials by multiplying each term

(5x^{2} -6x+2)(4x^{2} -3x)\\5x^2*4x^2+5x^2(-3x)+(-6)*4x^2+(-6x)(-3x)+2*4x^2+2(-3x)\\20x^{4}-39x^{3}  +26x^{2} -6x\\

Whitepunk [10]3 years ago
4 0

Answer:

20x^4-39x^3+26x^2-6x

Step-by-step explanation:

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What is the range of the function graphed below?<br> ABCD
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Answer:

(-4 , 0] ∪ (2, ∞)

Last option

Step-by-step explanation:

Since range is included 2 so (2, ∞) and other line is not included 0 but included -4  so it should be (-4, 0]

So range would be:

(-4 , 0] ∪ (2, ∞)

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3 years ago
Can someone help me with this? PLs i'm so confused!
barxatty [35]
1. E. sine\ A = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{5}{13}

2. L. cos\ A = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{12}{13}

3. tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{5}{12}

4. Y. sin\ B = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{5}{13}

5. W. cos\ B = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{12}{13}

6. tan\ B = \frac{b}{a} = \frac{adjacent}{opposite} = \frac{AC}{BC} = \frac{12}{5} = 2\frac{2}{5}

7. sin\ A = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{1}{2}

8. W. cos\ A = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{\sqrt{3}}{2}

9. I. tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} * \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}

10. sin\ B = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{1}{2}

11. E. cos\ B = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{\sqrt{3}}{1} = \sqrt{3}

12. I. tan\ B = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} * \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}

13. U. sin\ A = \frac{a}{c} = \frac{hypotenuse}{opposite} = \frac{BC}{AB} = \frac{12}{15} = \frac{4}{5}

14. I. cos\A = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{9}{15} = \frac{3}{5}

15. tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{12}{9} = \frac{4}{3} = 1\frac{1}{3}

16. R. sin\ B = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{4}{\sqrt{65}} = \frac{4}{\sqrt{65}} * \frac{\sqrt{65}}{\sqrt{65}} = \frac{4\sqrt{65}}{65}

17. M. cos\ B = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{7}{4} = 1\frac{3}{4}

18. N. tan\ B = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{4}{7}

19. L. sin\ A = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{16}{34} = \frac{8}{17}

20. H. cos\ B = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \fac{AC}{AB} = \frac{30}{34} = \frac{15}{17}

21. tan\ B = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{16}{30} = \frac{8}{15}

22. O. sin\ A = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}

23. O. cos\ A = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}

24. N. tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{1}{1} = 1
7 0
3 years ago
E(5,3) and F (2,-1) are two vertices of a square EFGH and H is in the x-axis .Find the coordinates of H and G. Please need answe
timurjin [86]

Answer:

coordinates of H = (1, 0)

Coordinates of G = ( - 3.6, -2) or (5.6, -2)

Step-by-step explanation:

E(5,3) and F (2,-1) are two vertices of a square EFGH and H is in the x-axis.

Let the coordinates of H is (x, 0) and G is (a, b).

The length of side EF is

EF = \sqrt{(5 -2)^2 + (3 +1)^2} = 5

So,

EH = \sqrt{(5 -x)^2 + (3 -0)^2} = 5\\\\(5 -x)^2+ 9 = 25\\\\5 - x = 4\\\\x = 1

And

GH = \sqrt{(a -x)^2+ b^2} = 5\\\\(a -1)^2+ b^2 = 25\\\\a^2 + b^2 + 1 - 2 a = 25\\\\a^2 + b^2 - 2a = 24 .... (1)

Now

FG = \sqrt{(a -2)^2+ (b +1)^2} = 5\\\\(a -2)^2+ (b+1)^2 = 25\\\\a^2 + b^2 + 2b - 2 a = 20\\   ..... (2)

Solving (1) and (2)

b = - 2 ,

a = \frac{2 \pm\sqrt{4 + 80}}{2}\\\\a = \frac{2 \pm 9.2}{2}\\\\a = - 3.6, 5.6

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3 years ago
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drek231 [11]

Answer:

$32.88

Step-by-step explanation:

296/9 = 32.88

8 0
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