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VashaNatasha [74]
3 years ago
14

Six years ago my son was one-third my age at that time.

Mathematics
1 answer:
Vika [28.1K]3 years ago
4 0

m - my age

s - son's age

s-6=\dfrac{m-6}{3}\\s+6=\dfrac{m+6}{2}\\\\3s-18=m-6\\2s+12=m+6\\\\m=3s-12\\m=2s+6\\\\3s-12=2s+6\\s=18

<u>He's 18</u>

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Solve the equation using square roots. Write your answer in simplest form.<br> Question in picture
GalinKa [24]

Answer:

The solutions to the quadratic equation involving square roots are:

x=2\sqrt{7}i+8,\:x=-2\sqrt{7}i+8

Step-by-step explanation:

Given the equation

\frac{1}{4}\left(x-8\right)^2+8=1

subtract 8 from both sides

\frac{1}{4}\left(x-8\right)^2+8-8=1-8

\frac{1}{4}\left(x-8\right)^2=-7

\left(x-8\right)^2=-28

\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)}

solving

x-8=\sqrt{-28}

x-8=\sqrt{-1}\sqrt{28}

x-8=\sqrt{28}i       ∵ \sqrt{-1}=i

x=2\sqrt{7}i+8

similarly solving

x-8=-\sqrt{-28}

x-8=-2\sqrt{7}i

x=-2\sqrt{7}i+8

Therefore, the solutions to the quadratic equation involving square roots are:

x=2\sqrt{7}i+8,\:x=-2\sqrt{7}i+8

5 0
3 years ago
The side length of a square can be represented by the expression 2x − 5. Write a polynomial that represents the area of the squa
Helen [10]

Answer:

Step-by-step explanation:

area of square A=(2x-5)²

5 0
4 years ago
If f(x)=4x+2<br><br>f(-4)<br><br>can you please explain with steps​
kompoz [17]

Answer:

replace x with -4

f(-4)= 4(-4) + 2

4(-4) + 2 = -16 + 2 = -14

6 0
4 years ago
Help please thanks so much
UNO [17]

Answer:

d

Step-by-step explanation:

7 0
3 years ago
Pls pls pls help if ur good with angles and sides ​
liq [111]

Answer:

∆PQR: Equilateral Triangle

∆PRT: Scalene Triangle

∆TQS: Isosceles Triangle

∆QNP: Right Triangle

Step-by-step explanation:

∆PQR is an Equilateral Triangle because all of its angles are equal (60°).

∆PRT is a Scalene Triangle because none of its sides are equal.

The triangle above it is equilateral, and one side length is 14, meaning all of its side lengths are 14.

This makes the top of ∆PRT 14. The hypotenuse is given and it’s 18.

The last side left is clearly shorter than both, meaning none of the sides are equal.

∆TQS is an Isosceles Triangle because 2 angles are equal.

The base angles are both 76°.

∆QNP is a Right Triangle because segment QN and PR are perpendicular, creating two 90° angles on both sides.

For number 10, I suggest using a graphing calculator or desmos to graph the points. (Or you can just add the points but I suggest graphing).

From there, you can count the length of each side.

Segment DE: 9

Segment EF: 8

Segment FD: 10

This makes ∆DEF in number 10 a Scalene Triangle because none of its sides are equal.

For number 11, I also suggest graphing the points and counting the sides.

Segment DE: 7

Segment EF: 9

Segment FD: 7

This makes ∆DEF in number 11 an Isosceles Triangle because 2 of its sides are equal.

Hope this helps! :)

4 0
3 years ago
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