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3 years ago

10

Consider the following formula

1 answer:

Gelneren [198K]3 years ago

3 0

Hello
This question requires us to change the subject of a formula. This can be achieved by following the order of operations in reverse. First, isolate the terms with our variable of interest, x:
ax - bx = z - y
Then, we take x out as it is being multiplied to both a and b:
x(a - b) = z - y
Dividing (a - b) on both sides, we get:
x = (z - y) / (a - b)
Thus, the answer is A

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Rename the number 120,000= ? Ten thousands

galben [10]

12,000 is that what you mean .

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4 years ago

Evaluate 1/4 plus 5/8. please show work

adelina 88 [10]

Answer:

7/8.

Step-by-step explanation:

The LCD of 4 and 8 is 8.

So we convert: 1/4 = 2/8.

2/8 + 5/8 = 7/8 (answer).

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4 years ago

Identify the polygon with vertices A(5,0), B(2,4), C(−2,1), and D(1,−3), and then find the perimeter and area of the polygon. HE

inn [45]

Answer:

Part 1) The polygon is a square

Part 2) The perimeter is equal to $20\ units$

Part 3) The area is equal to $25\ units^{2}$

Step-by-step explanation:

we have

$A(5,0), B(2,4), C(-2,1),D(1,-3)$

Plot the points

see the attached figure

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance AB

$A(5,0),B(2,4)$

substitute in the formula

$d=\sqrt{(4-0)^{2}+(2-5)^{2}}$

$d=\sqrt{(4)^{2}+(-3)^{2}}$

$d=\sqrt{25}$

$AB=5\ units$

Find the distance BC

$B(2,4), C(-2,1)$

substitute in the formula

$d=\sqrt{(1-4)^{2}+(-2-2)^{2}}$

$d=\sqrt{(-3)^{2}+(-4)^{2}}$

$d=\sqrt{25}$

$BC=5\ units$

Find the distance CD

$C(-2,1),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-1)^{2}+(1+2)^{2}}$

$d=\sqrt{(-4)^{2}+(3)^{2}}$

$d=\sqrt{25}$

$CD=5\ units$

Find the distance AD

$A(5,0),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-0)^{2}+(1-5)^{2}}$

$d=\sqrt{(-3)^{2}+(-4)^{2}}$

$d=\sqrt{25}$

$AD=5\ units$

we have that

AB=BC=CD=AD

Find the distance BD (diagonal)

$B(2,4),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-4)^{2}+(1-2)^{2}}$

$d=\sqrt{(-7)^{2}+(-1)^{2}}$

$BD=\sqrt{50}\ units$

<em>Verify if the polygon is a square</em>

If the triangle BDA is a right triangle, then the polygon is a square

Applying the Pythagoras theorem

$BD^{2}=AD^{2}+AB^{2}$

substitute

$(\sqrt{50})^{2}=5^{2}+5^{2}$

$50=50$ -----> is true

so

The triangle BDA is a right triangle

therefore

The polygon is a square

<em>Find the Area of the polygon</em>

The area of a square is equal to

$A=b^{2}$

we have

$b=5\ units$

$A=5^{2}=25\ units^{2}$

<em>Find the perimeter of the polygon</em>

The perimeter of a square is equal to

$P=4b$

we have

$b=5\ units$

$P=4(5)=20\ units$

6 0

3 years ago

Please help me out with this question please...

anastassius [24]

I believe it’s the second option because there are two 3’s with different outputs

3 0

3 years ago

Read 2 more answers

If f(x) = 3x^2-10x+2 find the vertex of y=f(x+4)-5

allochka39001 [22]

$3 {(4 + x)}^{2} - 10(x + 4) + 2 - 5$
so
$3( {x}^{2} + 8x + 16) - 10x - 40 + 3$
so
$3 {x}^{2} + 14x + 11$
take differential calculus
$6x + 14 = 0$
you will find x then y

3 0

4 years ago