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Volgvan
3 years ago
5

the door frame has measurements of 3 feet by 8 feet. What is the length of the largest table that can be brought in the house on

a diagonal? someone please help me out !!!!!
Mathematics
1 answer:
goldenfox [79]3 years ago
7 0

8 ft. I’m not really sure though sorry

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16+2b-7b subtracting like terms
White raven [17]
It equals 16+5b
2b-7b= 5b
3 0
4 years ago
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Please please please help!! i need this asap and i have no where else to go!
nadezda [96]
1   1   1        2
2  -3   1      -11            -2R1 + R2 → R2
-1  2  -1        8                R1 + R3 → R3

1    1    1      2
0    -5  -1   -15             R2 ⇔ R3
0    3    0     10

1    1    1      2
0    3    0     10             -R3
0   -5   -1    -15

1    1    1      2
0    3    0     10             1/3 R2
0    5    1     15

1    1    1      2                   -R2 + R1
0    1    0     10/3             -5R2 + R3
0    5    1     15

1    0    1      -4/3              
0    1    0     10/3             -R3 + R1
0    0    1      -5/3

1    0    0       1/3             
0    1    0     10/3           
0    0    1      -5/3

Therefore, x = 1/3, y = 10/3, z = -5/3
7 0
3 years ago
Read 2 more answers
What is 79.35 divided by 5
liraira [26]
79.35 divided by 5 equals: <u><em>15.87</em></u>
<em>~~~~~~~~~~~~~~~~~~~~~~~~~~~</em>
To check:
    243
    15.87
<span><u>  x  5.00</u>
</span>     0000
<u>    0000</u>
 79.35
<em>~~~~~~~~~~~~~~~~~~~~~~~~~~~~</em>
Hope this helped! :)
<span><em>Any more questions, please ask! :)</em></span>


5 0
4 years ago
The Sunday newspaper costs $2.10 issue delivered and the Saturday and weekday papers cost 40 cents per issue delivered how many
In-s [12.5K]

26 weeks is what I got

6 0
3 years ago
The radius of a sphere is measured to be 3.0 inches. If the measurement is correct within 0.01 inches, use differentials to esti
Zarrin [17]

Answer:

ΔV = 0.36π   in³

Step-by-step explanation:

Given that:

The radius of a sphere = 3.0

If the measurement is correct within 0.01 inches

i.e the change in the radius Δr = 0.01

The objective is to use differentials to estimate the error in the volume of sphere.

We all know that the volume of a sphere

V =  \dfrac{4}{3} \pi r^3

The differential of V with respect to r is:

\dfrac{dV}{dr }= 4 \pi r^2

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which can be re-written as:

ΔV = 4 πr² Δr

ΔV = 4 × π × (3)² × 0.01

ΔV = 0.36π   in³

5 0
3 years ago
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