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Ksenya-84 [330]
3 years ago
15

Use diagram to answer 3(x+6)=30

Mathematics
2 answers:
Alja [10]3 years ago
6 0
The answer is 4
Plug in method

3 (4)+6 =30 plug in value for x
3(4)+3(6) =30 distributive prop.
12+18=30 equals 30
Katyanochek1 [597]3 years ago
5 0

Answer:

Step-by-step explanation:

3x + 18 = 30

3x + 18 -18 = 30 - 18

3x = 12

x = 4

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Mrs.joshi bought a saree for Rs 1750.she sold it at a profit of 4%.what would be her profit or loss percent ?​
bekas [8.4K]

Answer: Rs 1820

Step-by-step explanation:

This is profit, thus is it a percentage increase of 4%. Thus, she sold the saree for 104% of what she bought it for, Rs 1750.  Thus, simply do 1.04*1750 to get 1820.

<em>Hope it helps <3</em>

7 0
3 years ago
The scale drawing of a rectangular room measures 10cm long by 6cm wide. The actual width of the room is 9 metres. What is the ac
Snezhnost [94]

Answer:

48 meters

Step-by-step explanation:

The first thing is to calculate the scale, since we know both widths it is possible:

we pass the 9 meters to centimeters = 900 cm

900/6 = 150

So the scale would be 1: 150

Which means that the length is 150 times bigger.

10 * 150 = 1500cm or what equals 15m

the perimeter is the sum of all the sides, therefore:

P = 15 + 15 + 9 + 9

P = 48

Which means that the perimeter of the room is 48 meters

8 0
3 years ago
Find all solutions to the following quadratic equations, and write each equation in factored form.
dexar [7]

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

\left(2x-1\right)^2=\quad 4x^2-4x+1\\\left(x+1\right)^2-3=\quad x^2+2x-2\\\\4x^2-4x+1=x^2+2x-2\\4x^2-4x+1+2=x^2+2x-2+2\\4x^2-4x+3=x^2+2x\\4x^2-4x+3-2x=x^2+2x-2x\\4x^2-6x+3=x^2\\4x^2-6x+3-x^2=x^2-x^2\\3x^2-6x+3=0

\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

3 0
3 years ago
What is 2/3 divided b 1/6 =?
Xelga [282]

Answer:

4

Step-by-step explanation:

\frac{2}{3} \div \frac{1}{6} can also be written as  \frac{2}{3} \times \frac{6}{1}. Evaluating this gives \frac{2 \times 6}{1 \times 3} = \frac{12}{3} = 4

5 0
2 years ago
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