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3 years ago

Please help me with this problem.

Mathematics

1 answer:

Alik [6]3 years ago

8 0

We're going to use the equation given but let's first make sure we understand what the equation IS saying:

Area of a kite= [(1st diagonal) × (2nd diagonal)]/2

Now, let's find the 2 diagonals. In the case of kite shapes one of the diagonals is the line that goes from W to Y and the other is the line going from X to Z. In a kite they will always cross (no matter their lengths).

Now that we know the diagonals we can add the lengths that are given for each line to know what to multiply.

Line W to Y: 20+12= 32

Line X to Z: 12+12= 24

Now let's plug these numbers into the equation (it doesn't matter the order of the diagonals because we're just gonna be multiplying them either way!)

So...

(32×24)/2= 384 units squared (because it is area)

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3. Sophie and Jackie each have a collection of baseball cards, Jackie has 5 more cards than

IRISSAK [1]

Answer:

Sophie has 12.5 cards

Step-by-step explanation:

Let

x -----> number of cards Sophie has

y -----> number of cards Jackie has

we know that

x+y=30 -----> equation A

y=x+5 -----> equation B

Solve by substitution

Substitute equation B in equation A and solve for x

x+(x+5)=30

2x=30-5

x=12.5 cards

Note I assume the problem was invented without taking into account the result, because the amount of cards should be a whole number

5 0

2 years ago

Need help for 18) 21) Solve each equation and check. Show all work please

Alexxandr [17]

18) (10)7d/10=35(10)
7d/7=350/7
d=50
21) 3/4 w=27
(4/3)3/4 w=27(4/3)
w=36
Hope this helps!

5 0

3 years ago

Someone please be awesome and help me please :(

solong [7]

Answer:

$(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0$

$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$

$x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}$

Step-by-step explanation:

$x^2+\frac{b}{a}x+\frac{c}{a}=0$

They wanted to complete the square so they took the thing in front of x and divided by 2 then squared. Whatever you add in, you must take out.

$x^2+\frac{b}{a}x+(\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0$

Now we are read to write that one part (the first three terms together) as a square:

$(x+\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0$

I don't see this but what happens if we find a common denominator for those 2 terms after the square. (b/2a)^2=b^2/4a^2 so we need to multiply that one fraction by 4a/4a.

$(x+\frac{b}{2a})^2+\frac{4ac}{4a^2}-\frac{b^2}{4a^2}=0$