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3 years ago

5

PLEASE HELP!! I’ll mark brainliest!! Image of the figure will be attached. 2.5 2.5 and 5 are the numbers! To get brainliest plea

se leave work attached or explanation!!! Part c
Jackson has a second cube identical to the first. He cuts off the triangular prism as shown below and throws
it away.
What is the volume, in cubic centimeters, of the remaining solid? Show or explain all your work
Work:

2 answers:

Harman [31]3 years ago

6 0

Answer:

$\Large \boxed{\sf 109.375 \ cm^3}$

Step-by-step explanation:

Volume of remaining solid = Volume of cube - Volume of triangular prism

Volume of cube:

$5^3=125$

Volume of triangular prism:

$2.5 \times 2.5 \times 0.5 \times 5=15.625$

Volume of remaining solid:

$125-15.625=109.375$

balandron [24]3 years ago

5 0

Answer:

109.375 cm³

Step-by-step explanation:

<u>The volume of the cube is:</u>

V = a³ = 5³ = 125 cm³

<u>The volume of the cut prism:</u>

V = Bh, B- area of the base, h- height

The base is a right triangle with legs of 2.5 each.

<u>The volume is:</u>

V = 1/2*2.5*2.5*5 = 15.625 cm³

<u>Required volume of the solid is the difference of the above volumes:</u>

125 - 15.625 = 109.375 cm³

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Find all solutions to the following quadratic equations, and write each equation in factored form.

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Answer:

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

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

(c) The solutions are: $x=i-2,\:x=-i-2$

(d) The solutions are: $x=-\frac{3}{2}+i\frac{\sqrt{7}}{2},\:x=-\frac{3}{2}-i\frac{\sqrt{7}}{2}$

(e) The solutions are: $x=1,\:x=-1,\:x=\sqrt{5}i,\:x=-\sqrt{5}i$

(f) The solutions are: $x=1$

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

(h) The solutions are: $x=2,\:x=2i,\:x=-2i$

Step-by-step explanation:

To find the solutions of these quadratic equations you must:

(a) For $x^2+25=0$

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

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

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

$\mathrm{Simplify}\:\sqrt{-25}\\\\\mathrm{Apply\:radical\:rule}:\quad \sqrt{-a}=\sqrt{-1}\sqrt{a}\\\\\sqrt{-25}=\sqrt{-1}\sqrt{25}\\\\\mathrm{Apply\:imaginary\:number\:rule}:\quad \sqrt{-1}=i\\\\\sqrt{-25}=\sqrt{25}i\\\\\sqrt{-25}=5i$

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

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

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

The solutions are: $x=i-2,\:x=-i-2$

(d) For $\left(x+2\right)^2=x$

$\mathrm{Expand\:}\left(x+2\right)^2= x^2+4x+4$

$x^2+4x+4=x\\x^2+4x+4-x=x-x\\x^2+3x+4=0$

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are:

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=3,\:c=4:\quad x_{1,\:2}=\frac{-3\pm \sqrt{3^2-4\cdot \:1\cdot \:4}}{2\cdot \:1}$

$x_1=\frac{-3+\sqrt{3^2-4\cdot \:1\cdot \:4}}{2\cdot \:1}=\quad -\frac{3}{2}+i\frac{\sqrt{7}}{2}\\\\x_2=\frac{-3-\sqrt{3^2-4\cdot \:1\cdot \:4}}{2\cdot \:1}=\quad -\frac{3}{2}-i\frac{\sqrt{7}}{2}$

The solutions are: $x=-\frac{3}{2}+i\frac{\sqrt{7}}{2},\:x=-\frac{3}{2}-i\frac{\sqrt{7}}{2}$

(e) For $\left(x^2+1\right)^2+2\left(x^2+1\right)-8=0$

$\left(x^2+1\right)^2= x^4+2x^2+1\\\\2\left(x^2+1\right)= 2x^2+2\\\\x^4+2x^2+1+2x^2+2-8\\x^4+4x^2-5$

$\mathrm{Rewrite\:the\:equation\:with\:}u=x^2\mathrm{\:and\:}u^2=x^4\\u^2+4u-5=0\\\\\mathrm{Solve\:with\:the\:quadratic\:equation}\:u^2+4u-5=0$

$u_1=\frac{-4+\sqrt{4^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}=\quad 1\\\\u_2=\frac{-4-\sqrt{4^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}=\quad -5$

$\mathrm{Substitute\:back}\:u=x^2,\:\mathrm{solve\:for}\:x\\\\\mathrm{Solve\:}\:x^2=1=\quad x=1,\:x=-1\\\\\mathrm{Solve\:}\:x^2=-5=\quad x=\sqrt{5}i,\:x=-\sqrt{5}i$

The solutions are: $x=1,\:x=-1,\:x=\sqrt{5}i,\:x=-\sqrt{5}i$

(f) For $\left(2x-1\right)^2=\left(x+1\right)^2-3$

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$\mathrm{For\:}\quad a=3,\:b=-6,\:c=3:\quad x_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:3\cdot \:3}}{2\cdot \:3}\\\\x_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{0}}{2\cdot \:3}\\x=\frac{-\left(-6\right)}{2\cdot \:3}\\x=1$

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 $x^3-2x^2+4x-8=0$

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

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

$x-2=0:\quad x=2\\x^2+4=0:\quad x=2i,\:x=-2i$

The solutions are: $x=2,\:x=2i,\:x=-2i$

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3 years ago

Please help me. .....

wlad13 [49]

Answer:

24 pounds

Step-by-step explanation:

Step 1: Find out how many pens he bought

12 pens per pack

Charles bought 60 packs

60 packs x 12 pens = 720 pens

Step 2: Find out how much Charles spent

2.80 per pack

Charles bought 60 packs

60 x 2.80 = 168 pounds

Step 3: Find out how many pens Charles sold

Charles sold 2/3 of his pens

2/3 x 720 = 480

Step 4: Find out how manypens he sold

Charles sold the pens for 40p

40p x 480 = 19200p

Step 5: Convert pences to pounds

19200p = 192 pounds

Step 6: Find the profit

192 - 168 = 24 pounds

Step 7: Therefore statement

Therefore he mad 24 pounds of profit

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3 years ago