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2 years ago

Please help and show work if you can

Mathematics

1 answer:

UkoKoshka [18]2 years ago

5 0

Answer:

38. <1 = $38^{o}$ , <2 = $32^{o}$ , <3 = $110^{o}$

39. <1 = $71^{o}$ , <2 = $28^{o}$ , and <3 = $81^{o}$

Step-by-step explanation:

38. Sum of angles in a triangle = $180^{o}$

So that;

38 + 110 + <2 = $180^{o}$

148 + <2 = $180^{o}$

<2 = $180^{o}$ - 148

<2 = $32^{o}$

Thus,

<1 = $38^{o}$ (alternate angle principle)

<3 = $110^{o}$

Therefore;

<1 = $38^{o}$ , <2 = $32^{o}$ , <3 = $110^{o}$

39. <2 = $28^{o}$ (alternate angle principle)

So that;

sum of angles in a triangle = $180^{o}$

<1 + <2 + 81 = $180^{o}$

<1 + $28^{o}$ + 81 = $180^{o}$

<1 + 109 = $180^{o}$

<1 = $180^{o}$ - 109

= $71^{o}$

<1 = $71^{o}$

<3 = $81^{o}$

<1 = $71^{o}$ , <2 = $28^{o}$ , and <3 = $81^{o}$

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(a) The solutions are: $x=5i,\:x=-5i$

(b) The solutions are: $x=3i,\:x=-3i$

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Step-by-step explanation:

To find the solutions of these quadratic equations you must:

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$\mathrm{Subtract\:}25\mathrm{\:from\:both\:sides}\\x^2+25-25=0-25$

$\mathrm{Simplify}\\x^2=-25$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{-25},\:x=-\sqrt{-25}$

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$-\sqrt{-25}=-5i$

The solutions are: $x=5i,\:x=-5i$

(b) For $-x^2-16=-7$

$-x^2-16+16=-7+16\\-x^2=9\\\frac{-x^2}{-1}=\frac{9}{-1}\\x^2=-9\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\x=\sqrt{-9},\:x=-\sqrt{-9}$

The solutions are: $x=3i,\:x=-3i$

(c) For $\left(x+2\right)^2+1=0$

$\left(x+2\right)^2+1-1=0-1\\\left(x+2\right)^2=-1\\\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x+2=\sqrt{-1}\\x+2=i\\x=i-2\\\\x+2=-\sqrt{-1}\\x+2=-i\\x=-i-2$

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The solutions are: $x=1$

(g) For $x^3+x^2-2x=0$

$x^3+x^2-2x=x\left(x^2+x-2\right)\\\\x^2+x-2:\quad \left(x-1\right)\left(x+2\right)\\\\x^3+x^2-2x=x\left(x-1\right)\left(x+2\right)=0$

Using the Zero Factor Theorem: = 0 if and only if = 0 or = 0

$x=0\\x-1=0:\quad x=1\\x+2=0:\quad x=-2$

The solutions are: $x=0,\:x=1,\:x=-2$

(h) For $x^3-2x^2+4x-8=0$

$x^3-2x^2+4x-8=\left(x^3-2x^2\right)+\left(4x-8\right)\\x^3-2x^2+4x-8=x^2\left(x-2\right)+4\left(x-2\right)\\x^3-2x^2+4x-8=\left(x-2\right)\left(x^2+4\right)$

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2 years ago

Which numbers are 9 units from −5 on this number line?

ExtremeBDS [4]

Using the number line, the numbers that are 9 units from -5 are: -14 and 4.

<h3>How to Locate a Number on a Number line?</h3>

To find two numbers that cover the same units from a given point on a number line, we can simply do the following:

Count the number of units given backwards/to the left from the point stated to get the first number.
Count the number of units given forwards/to the right from the point stated to get the second number.

Thus, we are asked to find the numbers that would be 9 units from -5, using the number line.

Count 9 units backwards/to the left from -5 to get the first number, which is: -14

Count 9 units forwards/to the right from the -5 to get the second number, which is: 4.

Therefore, the numbers that are 9 units from -5 on the number line, are: -14 and 4.

Learn more about number line on:

brainly.com/question/24644930

#SPJ1

8 0

1 year ago

Please help and show work if you can​

Please help and show work if you can