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3 years ago

Question:

Mathematics

1 answer:

kaheart [24]3 years ago

6 0

Answer:

-29

Step-by-step explanation:

${\sqrt{7 - y }}^{2} = {6}^{2}$

$7 - y = {6}^{2}$

y = 7-36

y = -29

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Matt and Alex are walking home from school one day and Alex says he knows a short cut. Matt says that the normal way they go is

nataly862011 [7]

Answer:mat

Step-by-step explanation:300 is less then 400

3 0

2 years ago

Read 2 more answers

Use elimination method to solve the silmutaneous equations 2x +3y=1 3x=2y+8 (4marks)

kotegsom [21]

Answer:

2x +3y=1⇒equation 1

3x=2y+8

3x-2y=8⇒equation 2

by elimination method,

2x+3y=1 ×2

3x-2y=8 ×3

4x+6y=2

9x-6y=24

cut 6y and -6y

4x+6y=2

9x-6y=24

---------------

13x+0=26

13x=26

x=26/13

x=2

substitute x=2 in equation 1

2(2)+3y=1

4+3y=1

3y=1-4

3y=-3

y=-3/3

y=-1

therefore, x=2 and y=-1

hope it helps u!

6 0

2 years ago

Help me i need help badly again

Irina-Kira [14]

Alright, given that it's a male then

total of 4+6+2+2 or 14 pepole are male
4 of them are freshmen

so the probalbity would be 4/14 or 2/7 or aprox 0.28571428571428571428571428571429

about 29%

8 0

3 years ago

Which value is NOT a solution of 8x3 – 1 = 0?

Tpy6a [65]

Note: As you may have unintentionally missed to add the value choices. But, I would make sure to explain the concept so that you may improve your understanding in terms of solving these type of questions.

Answer:

Any value other than the values $x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}$ will not be a solution of $8x^3\:-\:1\:=\:0$ .

Step-by-step explanation:

Considering the equation

$8x^3\:-\:1\:=\:0$

Steps to solve the equation

$8x^3-1=0$

$\mathrm{Add\:}1\mathrm{\:to\:both\:sides}$

$8x^3-1+1=0+1$

$\mathrm{Simplify}$

$x^3=\frac{1}{8}$

$\mathrm{Divide\:both\:sides\:by\:}8$

$\frac{8x^3}{8}=\frac{1}{8}$

$\mathrm{Simplify}$

$x^3=\frac{1}{8}$

$\mathrm{For\:}x^3=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}$

$x=\sqrt[3]{\frac{1}{8}},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1+\sqrt{3}i}{2},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1-\sqrt{3}i}{2}$

So,

$x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}$

Therefore,

Any value other than the values $x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}$ will not be a solution of $8x^3\:-\:1\:=\:0$ .

Keywords: solution, value

Learn more about equation solution from brainly.com/question/1679491

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7 0

2 years ago

Solve easy equation (for points) 8+x+15=20

Andru [333]

Answer:

x= -3

Step-by-step explanation:

You're Welcome :)

7 0

3 years ago