All categories

3 years ago

3.0

Mathematics

1 answer:

Alex73 [517]3 years ago

3 0

Answer:

Solution given:

AD=1.8

AC=3.0

we have

∆ADC is a right angled triangle

Cos A=b/h

<A=cos-¹ (1.8/3)

<A=53.13°

again

∆ABC is a right angled triangle

Tan A=p/b=BC/AC

Tan 53.13×3=BC

BC=4

NOW

Area of triangle =½(AC*BC)=½(4×3)=6 square units is a required area.

You might be interested in

The Chinese system presented in the text has symbols for 1 through 9, and also for 10, 100, and 1000. What is the greatest num

Naddika [18.5K]

To get the biggest possible number, use all the symbols at once. 1000+100+10+9+8+7+6+5+4+3+2+1=what?

4 0

3 years ago

What is equivalent to the equation to 1/4 (8 + 6z + 12) ?

Fed [463]

1/4 (8 + 6z + 12)

------------------------------------------------------------
Combine like terms :
------------------------------------------------------------
1/4 (6z + 20)

------------------------------------------------------------
Apply distributive property :
------------------------------------------------------------
1/4(6z) + 1/4(20)
3/2z + 5

------------------------------------------------------------
Answer: 3/2z + 5
------------------------------------------------------------

7 0

3 years ago

25% of American households have only dogs (one or more dogs) 15% of American households have only cats (one or more cats) 10% of

sergeinik [125]

Answer:

a) P=0.2503

b) P=0.2759

c) P=0.3874

d) P=0.2051

Step-by-step explanation:

We have this information:

25% of American households have only dogs (one or more dogs)

15% of American households have only cats (one or more cats)

10% of American households have dogs and cats (one or more of each)

50% of American households do not have any dogs or cats.

The sample is n=10

a) Probability that exactly 3 have only dogs (p=0.25)

$P(x=3)=\binom{10}{3}0.25^30.75^7=120*0.01563*0.13348=0.25028$

b) Probability that exactly 2 has only cats (p=0.15)

$P(x=2)=\binom{10}{2}0.15^20.85^8=45*0.0225*0.27249=0.2759$

c) Probability that exactly 1 has cats and dogs (p=0.1)

$P(x=1)=\binom{10}{1}0.10^10.90^0=10*0.1*0.38742=0.38742$

d) Probability that exactly 4 has neither cats or dogs (p=0.5)

$P(x=4)=\binom{10}{4}0.50^40.50^6=210*0.0625*0.01563=0.20508$

8 0

3 years ago

Sam, Jimmy, and Andre shared a number of pennies in the ratio 8:5:3. Sam and Andre received a

ddd [48]

Jimmy received 122 pennies more than Andre

Solution:

Given that , Sam, Jimmy, and Andre shared a number of pennies in the ratio 8:5:3

Sam and Andre received a total of 671 pennies.

Now, let the H.C.F of number of pennies with them be "x"

Then, number of pennies with Sam will be 8x

Number of pennies with Jimmy will be 5x

Number of pennies with Andre will be 3x

Now, we know that, Sam and Andre has a total of 671 pennies

$\begin{array}{l}{\Rightarrow \quad 8 x+3 x=671} \\\\ {\Rightarrow \quad 11 x=671} \\\\ {\Rightarrow \quad x=61}\end{array}$

$\begin{array}{l}{\text { Then, Sam will have } 8 \times 61=488 \text { pennies, }} \\\\ {\text { Jimmy will have } 5 \times 61=305 \text { pennies, }} \\\\ {\text { Andre will have } 3 \times 61=183 \text { pennies. }}\end{array}$

Calculating how many more pennies Jimmy received than Andre:

Number of Pennies Jimmy received more than Andre = Pennies with jimmy - Pennies with Andre

= 305 – 183 = 122

Hence, jimmy has 122 pennies more than Andre

7 0

3 years ago

Find all the zeroes of the equation. Show all your work. -3x^4+27x^2+1200=0

madam [21]

There are two things you should do right away

Divide through by minus 3

Substitute another letter (say z) for x^2

Step One

-3x^4/-3 +27x^2/-3 + 1200/-3 = 0/-3

x^4 - 9x^2 - 400 = 0

Step 2

Let z = x^2

(x^2)^2 - 9(x^2) - 400 = 0

z^2 - 9z^2 - 400 = 0

Step 3

By factoring solve for z

(z - 25)(z + 16) = 0

z = 25

x^2 = 25

x = +/- 5

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

z + 16 = 0

z = - 16

x^2 = - 16

x = sqrt(-16)

x = +/- 4i

I've included a graph to show that the zeros are correct.

4 0

3 years ago