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2 years ago

14

∆ADC is formed by reflecting ∆ABC across line segment , as shown in the figure. If the length of is 4 units, the area of ∆ADC is

________ square units.

2 answers:

Stolb23 [73]2 years ago

4 0

Answer: the area of ∆ADC is

<u><em>6 </em></u>

square units.

koban [17]2 years ago

3 0

Answer - 6 sq.units

Solution - As the triangle ADC is the reflection of the triangle ABC, so the area of both the triangles will be same.

Here BE is the height and AC is the base of ABC.

Area of triangle ABC = $\frac{1}{2}$ ×Base×Height

= $\frac{1}{2}$ ×3×4

= 6 sq.units

So the area of triangle ADC = 6 sq.units

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Ratio for 669 and 221

alexandr402 [8]

Hello from MrBillDoesMath!

Answer:

669/221

Discussion:

Ratio of 669 to 221 = 669/221.

Now, 669 = 3 * 223 and 221 = 13 * 17. Note that all numbers involved in the factorizations are distinct primes so nothing cancels out in their ratio. That is,

669/221 can not be reduced further.

Thank you,

MrB

5 0

2 years ago

Read 2 more answers

Which equation is y = 3(x – 2)2 – (x – 5)2 rewritten in vertex form? Y = 3 (x minus seven-halves) squared minus StartFraction 27

densk [106]

Answer: y = 2 (x minus one-half) squared minus StartFraction 27 Over 2 EndFraction

or

$y=2((x-\dfrac{1}{2})^2)-\dfrac{27}{2}$

Step-by-step explanation:

Vertex form of equation : $f (x) = a(x - h)^2 + k,$ where (h, k) is the vertex of the parabola.

$y=3(x-2)^2-(x-5)^2\\\\=3(x^2+4-4x)-(x^2+25-10x)\\\\=3x^2+12-12x-x^2-25+10x\\\\=2x^2-2x-13\\\\=2(x^2-x-\dfrac{13}{2})\\\\=2(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{13}{2})\\\\=2((x-\dfrac{1}{2})^2-\dfrac{1+26}{4})\\\\=2((x-\dfrac{1}{2})^2-\dfrac{27}{4})=2((x-\dfrac{1}{2})^2)-\dfrac{27}{2}$

Hence, the vertex form of the equation is $y=2((x-\dfrac{1}{2})^2)-\dfrac{27}{2}$

8 0

2 years ago

Read 2 more answers

For the binomial expansion of (x + y)^10, the value of k in the term 210x 6y k is a) 6 b) 4 c) 5 d) 7

Sloan [31]

Answer:

a) 6

Step-by-step explanation:

Expanding the polynomial using the formula:

$(x+y)^n=\sum_{k=0}^n \binom{n}{k} x^{n-k} y^k$

Also

$\binom{n}{k}=\frac{n!}{(n-k)!k!}$

I think you mean $210x^6y^4$

We can deduce that this term will be located somewhere in the middle. So I will calculate $k= 5; k=6 \text{ and } k =7$ .

For $k=5$

$\binom{10}{5} (y)^{10-5} (x)^{5}=\frac{10!}{(10-5)! 5!}(y)^{5} (x)^{5}= \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5! }{5! \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 } \\ =\frac{30240}{120} =252 x^{5} y^{5}$

Note that we actually don't need to do all this process. There's no necessity to calculate the binomial, just $x^{n-k} y^k$

For $k=6$

$\binom{10}{6} \left(y\right)^{10-6} \left(x\right)^{6}=\frac{10!}{(10-6)! 6!}\left(y\right)^{4} \left(x\right)^{6}=210 x^{6} y^{4}$

5 0

2 years ago

A 16-inch candle is lit and burns at a constant rate of 0.8 inches per hour. Let t represent the number of hours since the candl

Tamiku [17]

Answer:

<h2>See the explanation.</h2>

Step-by-step explanation:

a.

The initial length of the candle is 16 inch. It also given that, it burns with a constant rate of 0.8 inch per hour.

After one hour since the candle was lit, the length of the candle will be (16 - 0.8) = 15.2 inch.

After two hour since the candle was lit, the length of the candle will be (15.2 - 0.8) = 14.4 inch. The length of the candle after two hours can also be represented by {16 - 2(0.8)}.

Hence, the length of the candle after t hours when it was lit can be represented by the function, $f(t) = 16 - 0.8t$ . $f(t) = 0$ at t = 20.

b.

The domain of the function is 0 to 20.

c.

The range is 0 to 16.

8 0

2 years ago

A rectangular room has a width of 5 yards and length of 6 yards. How many square-feet carpet will be required for carpeting the

Sindrei [870]

Answer:

Option B $270\ ft^{2}$

Step-by-step explanation:

we know that

The area of the rectangle room, is equal to

$A=LW$

where

L is the length

W is the width

we have

$L=6\ yd$

$W=5\ yd$

Remember that

1 yard= 3 feet

Convert yards to feet

$L=6*3=18\ ft$

$W=5*3=15\ ft$

The area is equal to

$A=(18)(15)=270\ ft^{2}$

4 0

2 years ago