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2 years ago

#9 Can anyone please help me, this is Pythagoras Theorem Converse.

Mathematics

1 answer:

77julia77 [94]2 years ago

3 0

Answer:

(D).

Step-by-step explanation:

$A_{b}$ = A + $A_{s}$

(A). 12 + 16 > 20

(B). 10 + 18 < 30

(C). 4 + 5 < 12

(D). 8 + 16 = 24

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Solve the equation 3x+5y=15 for y

larisa86 [58]

3x+5y = 15
5y = -3x + 15
y = -3/5(x) + 3

hope it helps

7 0

3 years ago

Help!! 50 points and brainliest!

Viktor [21]

Answer:

Second choice:

$x=2t$

$y=4t^2+4t-3$

Fifth choice:

$x=t+1$

$y=t^2+4t$

Step-by-step explanation:

Let's look at choice 1.

$x=t+1$

$y=t^2+2t$

I'm going to subtract 1 on both sides for the first equation giving me $x-1=t$ . I will replace the $t$ in the second equation with this substitution from equation 1.

$y=(x-1)^2+2(x-1)$

Expand using the distributive property and the identity $(u+v)^2=u^2+2uv+v^2$ :

$y=(x^2-2x+1)+(2x-2)$

$y=x^2+(-2x+2x)+(1-2)$

$y=x^2+0+-1$

$y=x^2$

So this not the desired result.

Let's look at choice 2.

$x=2t$

$y=4t^2+4t-3$

Solve the first equation for $t$ by dividing both sides by 2:

$t=\frac{x}{2}$ .

Let's plug this into equation 2:

$y=4(\frac{x}{2})^2+4(\frac{x}{2})-3$

$y=4(\frac{x^2}{4})+2x-3$

$y=x^2+2x-3$

This is the desired result.

Choice 3:

$x=t-3$

$y=t^2+2t$

Solve the first equation for $t$ by adding 3 on both sides:

$x+3=t$ .

Plug into second equation:

$y=(x+3)^2+2(x+3)$

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

$y=(x^2+6x+9)+(2x+6)$

$y=(x^2)+(6x+2x)+(9+6)$

$y=x^2+8x+15$

Not the desired result.

Choice 4:

$x=t^2$

$y=2t-3$

I'm going to solve the bottom equation for $t$ since I don't want to deal with square roots.

Add 3 on both sides:

$y+3=2t$

Divide both sides by 2:

$\frac{y+3}{2}=t$

Plug into equation 1:

$x=(\frac{y+3}{2})^2$

This is not the desired result because the $y$ variable will be squared now instead of the $x$ variable.

Choice 5:

$x=t+1$

$y=t^2+4t$

Solve the first equation for $t$ by subtracting 1 on both sides:

$x-1=t$ .

Plug into equation 2:

$y=(x-1)^2+4(x-1)$

Distribute and use the binomial square identity used earlier:

$y=(x^2-2x+1)+(4x-4)$

$y=(x^2)+(-2x+4x)+(1-4)$

$y=x^2+2x+-3$

$y=x^2+2x-3$ .

This is the desired result.

3 0

3 years ago

Read 2 more answers

27 + 5 x 6 A Little Problem With this so some help please?

Kay [80]

Answer:

Step-by-step explanation:

1) According to PEMDAS, multiply first: 27+ 5x6= 27+30

2) Then add: 27+30=57

3 0

3 years ago

Read 2 more answers

If the side lengths of a cube are 14 feet, what is the correct way to write the expression to represent the volume of the cube i

Kryger [21]

Answer: Choice A) $14^3$

This is the same as writing 14^3

===================================================

Explanation:

Since all the sides are 14 feet long, we can basically say length = width = height = 14.

Or put another way:

length = 14
width = 14
height = 14

The volume of the cube is length*width*height = 14*14*14 = 14^3 = 2744 cubic feet.

-----------

The term 14^3 has a base of 14 and exponent of 3.

The base is what we multiply out while the exponent tells us how many copies of the base to multiply. In this problem, we have 3 copies of 14 being multiplied. So that's why 14^3 = 14*14*14

If you were to say something like 3^14, then you would have 14 copies of 3 multiplied out, which is a much larger value. So we'll cross choice B off the list.

Choice C isn't correct either since again we're using exponents and not multiplication to tie together the 14 and 3.

Choice D would be correct if your teacher wanted it in expanded form, instead of exponent form.

6 0

2 years ago

Find the distance from point A to the given line: A(-1, 7), y=3x How do I solve this?

kipiarov [429]

he formula of a distance between a point (a, b)

and a line Ax + By + C = 0:

$d=\dfrac{|Aa+Bb+C|}{\sqrt{A^2+B^2}}$

We have the point A(-1, 7) → a = -1, b = 7

and the line y = 3x → 3x - y = 0 → A = 3, B = -1, C = 0.

Substitute:

$d=\dfrac{|(3)(-1)+(-1)(7)+0|}{\sqrt{3^2+(-1)^2}}=\dfrac{|-3-7|}{\sqrt{9+1}}=\dfrac{|-10|}{\sqrt{10}}=\dfrac{10\sqrt{10}}{\sqrt{10}\cdot\sqrt{10}}\\\\=\dfrac{10\sqrt{10}}{10}=\sqrt{10}$

<h3>Answer: d = √10</h3>

3 0

2 years ago