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Rudik [331]
3 years ago
10

Ebony is building a shelf to hold the two boxes shown. What is the least width she should make the shelf?

Mathematics
1 answer:
ki77a [65]3 years ago
7 0
There exists the same question from other source that shows the two boxes.

Box 1 has a length of 4/5 feet
Box 2 has a length of 3/4 feet

Because Ebony is trying to find the total amount of space needed to hold the boxes, use addition. 

Here, is the solution:
= 4 / 5 + 3 / 4
= (16 + 15) / 20
= 31 / 20

So, Ebony can make a shelf at least 31/20 feet wide or 1 11/20 feet.

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1) conspicuous consumption
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1. what is the perimeter of the triangle?<br> 2. what is the area of the triangle?
jonny [76]

Answer:

See below. A=45 p=39.64

Step-by-step explanation:

Hi = 9 squares (h)

Gh= 14 squares. (b)

Area of a triangle = 1/2 bxh

1/2 14 x 9

7x9= 45.

Area =45 squares.

Perimiter

A= side hi (9)

B= side gh (14)

C= side gi (unknown)

A^2 + b^2 = c^2

9^2 + 14^2 = 277

C^2 = 277

Square root of 277 =c

C length =16.64 squares

A+b+c = perimeter

9+14+16.64 = 39.64 squares.

Sorry I cannot see the units of measure in the photo

7 0
3 years ago
Points Z(4,3) X(-5,-2) Y(4,-2) plot them and then find the perimeter and area
mash [69]

Answer:

z(43)x(−5−2)y(4−2)

=−602xyz

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7 0
3 years ago
Simplify 4a²+1/3a+5+a²+4+1/3a
vladimir2022 [97]

Answer:

5a^2+\frac{2a}{3} + 9

Step-by-step explanation:

Refer to file. Trying something different. Should be easier to read than text format and is easier than mathML.

7 0
3 years ago
A sensor output was acquired for 32 seconds at a rate of 200 Hz and spectral analysis was performed using FFT. If the data set w
VashaNatasha [74]

Complete Question

A sensor output was acquired for 32 seconds at a rate of 200 Hz and spectral analysis was performed using FFT. If the data set was split into 5 segments (each 6.4 seconds long), what is the resulting:

a)F minimum

b) F maximum

c) Frequency resolution

Answer:

a) Fmax=100Hz

b) Fmin=0Hz

c) F_r=0.0313

Step-by-step explanation:

From the question we are told that:

Time t=32sec

FrequencyF=200Hz

Segments \mu=5

Generally the equation for Frequency Range is mathematically given by

Fmax=\frac{F}{2}

Fmax=100Hz

Therefore

a) Fmax=100Hz

b) Fmin=0Hz

c)

Generally the equation for Frequency Resolution is mathematically given by

F_r=\frac{F}{N}

Where

N=The Total dat points

N=Sampling Frequency *Time

N=200*32

N=6400

Therefore

F_r=\frac{200}{6400}

F_r=0.0313

6 0
3 years ago
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