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2 years ago

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Kite A B C D is shown. Lines are drawn from point A to point C and from point B to point D and intersect. In the kite, AC = 10 a

nd BD = 6. What is the area of kite ABCD? 15 square units 30 square units 45 square units 60 square units

1 answer:

Oduvanchick [21]2 years ago

4 0

Answer:

$30 u^{2}$

Step-by-step explanation:

The computation of the area of kite ABCD is shown below:

Given data

AC = 10 ;

BD = 6

As we can see from the attached figure that the Kite is a quadrilateral as it involves two adjacent sides i.e to be equal

Now the area of quadrilateral when the diagonals are given

So, it is

$\text { area of kite }=\frac{1}{2} \times d_{1} d_{2}$

where,

$d_{1}=10\ and\ d_{2}=6$

So, the area of the quadrilateral is

$=\frac{1}{2}(10)(6)\\\\=30 u^{2}$

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To solve this, we are going to use the compound interest formula: $A=P(1+ \frac{r}{n} )^{nt}$
where
$A$ is the final amount after $t$ years
$P$ is the initial investment
$r$ is the interest rate in decimal form
$n$ is the number of times the interest is compounded per year

For the first 4 years we know that: $P=2350$ , $r= \frac{4.2}{100} =0.042$ , $t=4$ , and since the problem is not specifying how often the interest is communed, we are going to assume it is compounded annually; therefore, $n=1$ . Lest replace those values in our formula:
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Now, for the next 6 years the intial investment will be the final amount from our previous step, so $P=2770.38$ . We also know that: $r= \frac{4.9}{100} =0.049$ , $t=6$ , and $n=1$ . Lets replace those values in our formula one more time:
$A=P(1+ \frac{r}{n} )^{nt}$
$A=2770.38(1+ \frac{0.049}{1})^{(1)(6)$
$A=2770.38(1+0.049)^6$
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Answer:

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Step-by-step explanation:

1. We can find this by suing combinations.

Here n= 10 and r= 3 so n C r

= 10 C 3= 120

2. E(X) = 8 and E(Y) = 3

Z = 2X - 3Y +5

E(Z ) = 2 E (X) - 3(E)(Y) +5 ( applying property for mean)

= 2(8) - 3(3)+ 5 = 16+5-9= 21-9= 12

V(X) = 9 and V(Y) = 6.

V(Z) = E(Z )²- V(X) *V(Y) (applying property for Variance for two variables )

= 144- 54= 90

3. 55 liters contain adulterated milk in 7: 4.

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the water will be 6/13 *55= 25.38 when rounded gives 25 liters of water

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if 1000 spherical of radius 1 cm are melted to form a bigger bulb. Then find the radius of bigger bulb

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Answer:

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Step-by-step explanation:

Given:

No. of small spherical bulb = 1,000

radius (r) of smaller bulbs = 1 cm

Required:

radius of the bigger bulb

SOLUTION:

The following equation represents the relationship of the volume of the smaller and bigger bulb,

$\frac{V_2}{V_1} = 1,000$

Where,

$V_2$ = volume of bigger bulb

$V_1$ = volume of smaller bulb

1,000 is the number of smaller bulbs melted to form the bigger bulb.

Volume of a sphere is given as, ⁴/3πr³

Therefore:

$V_2$ = ⁴/3*π*r³ = 4πr³/3

$V_1$ = ⁴/3*πr³ = ⁴/3*π*(1)³ = ⁴/3π*1 = 4π/3

Plug the above values into the equation below:

$\frac{V_2}{V_1} = 1,000$

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$\frac{4*pie*r^3}{3}*{\frac{3}{4*pie} = 1,000$

$\frac{4*pie*r^3*3}{3*4*pie} = 1,000$

$\frac{12*pie*r^3}{12*pie} = 1,000$

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$r = 10$ (taking the cube root of each side)

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