The drawing shows a plan for a living room.

HELPPPPPP math please only correct answer i beg you

Alekssandra [29.7K]

Answer:

the area is 15

Step-by-step explanation:

if you split this shape into a triangle and a square, the triangle would measure with a legnth og 4 and a with of 3, which = 12 and then you divide by 2 to = the triangles area(6) then you add tht to the area of the square 3*3=9. so the answer would be 15

8 0

2 years ago

In which number is the value of the 7 in the number 1.273

Ne4ueva [31]

The hundredths place

the 2 would be tenths, 1 is hundredths, and the 3 is thousandths

7 0

3 years ago

HELP ASAP HELP ASAP HELP ASAP HELP ADAP Find the values of m and n.

Shtirlitz [24]

M is 36° and N is 18°

6 0

3 years ago

Find the length of the height of the right trapezoid shown below, if it has the greatest possible area and its perimeter is equa

Artyom0805 [142]

Answer:

The height of the right trapezoid is $\frac{6}{5+\sqrt{3}}\ units$

Step-by-step explanation:

Let

x ----> the height of the right trapezoid in units

we know that

The perimeter of the figure is equal to

$P=AB+BC+CD+DH+HA$

we have

$P=6\ units$

$AB=BC=CH=HA=x$ ---> because is a square

substitute

$6=x+x+CD+DH+x$

$6=3x+CD+DH$ -----> equation A

In the right triangle CDH

$sin(30\°)=\frac{CH}{CD}$

$sin(30\°)=\frac{1}{2}$

so

Remember that $CH=x$

$\frac{1}{2}=\frac{x}{CD}$

$CD=2x$

$tan(30\°)=\frac{CH}{DH}$

$tan(30\°)=\frac{\sqrt{3}}{3}$

so

$\frac{\sqrt{3}}{3}=\frac{x}{DH}$

$DH=x\sqrt{3}$

substitute the values in the equation A

$6=3x+CD+DH$ -----> equation A

$CD=2x$

$DH=x\sqrt{3}$

$6=3x+2x+x\sqrt{3}$

$6=5x+x\sqrt{3}$

$6=x[5+\sqrt{3}]$

$x=\frac{6}{5+\sqrt{3}}\ units$

5 0

3 years ago

collect data (in decibels) of 10 different things in term of loudness. Arrange the data in ascending order & find mean, medi

ss7ja [257]

Whisper-20dB
Quiet Residence-30dB
Soft stereo in Residence-40dB
Average Speech-60dB
Cafeteria-80dB
Pneumatic Jackhammer-90dB
Loud crowd noise-100dB
Accelerating Bike-100dB
Rock concert-120dB
Jet Engine (75 ft away)-140dB

(1)Mode=100dB (since 100 is occurring the maximum no. of times, i.e. it has the highest frequency of 2)
(2)Mean=sum of observations/Total number of observations
=(20+30+40+60+80+90+100+100+120+140)/10
=770/10=10dB
(3)Median= 1/2[{n/2}th observation + {(n/2)+1}th observation], where, n=total no. of observations
So, 1/2[{10/2}th observation + {(10/2)+1}th observation]
=1/2[5th observation+6th observation]
=1/2[80+90] [Because:5th observation=80dB and 6th=90dB]
=1/2(170)
=85
Therefore, median=85dB

6 0

3 years ago