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2 years ago

A restaurant needs a block of ice that is exactly 480 cubic inches in volume.

Mathematics

2 answers:

Luda [366]2 years ago

7 0

Answer:

Part a: tape measure part b: 48 and part c: 2x24, 4x12, 8x6, and 3x16

Step-by-step explanation:

let me know if it’s correct

ivolga24 [154]2 years ago

4 0

Answer:

Step-by-step explanation:

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PLEASE RESPOND ASAP!!what is the value of z in the figure below??

Nookie1986 [14]

Answer:

126

Step-by-step explanation:

By cyclic quadrilateral theorem:

$(180 \degree - 85 \degree) = \frac{1}{2} (z \degree + 64 \degree) \\ 95 \degree \times 2 = z \degree + 64 \degree \\ 190 \degree = z \degree + 64 \degree \\ 190 \degree - 64 \degree = z \degree \\ 126\degree = z \degree \\ z = 126$

4 0

2 years ago

Please help with this square problem, the picture is shown.

Galina-37 [17]

Area of the upper rectangle = x(16 - 6) = 10x yd²

Area of the bottom rectangle = 6(x + 6) = 6x + 36 yd²

Total area = 10x + 6x + 36 = 16x + 36 yd²

5 0

3 years ago

Hello please help i’ll give brainliest

xenn [34]

Answer:

before "however"

Step-by-step explanation:

However is a throw-in word and should be both preceded and superseded by a comma.

3 0

2 years ago

Read 2 more answers

a tourist drove his car for the total of 17.5 hours while on a 3-day vacation on the first day he drove 4.5 hours in the second

AleksAgata [21]

Answer:

1.75 hours

Step-by-step explanation:

Time for day 1 : 4.5 hrs

Time for day 2: (4.5 ×2.5)

Time for day 3: let it be denoted by a variable ,say T.

Total Time= 17.5

(Time for day 1 ) + (Time for day 2) +(Time for day 3) =17.5

4.5 + (4.5×2.5) + T = 17.5

4.5+11.25+T =17.5

15.75+T=17.5

T=17.5-15.75

T= 1.75

7 0

2 years ago

(b) dy/dx = (x - y+ 1)^2

Elanso [62]

Substitute $v(x)=x-y(x)+1$ , so that

$\dfrac{\mathrm dv}{\mathrm dx}=1-\dfrac{\mathrm dy}{\mathrm dx}$

Then the resulting ODE in $v(x)$ is separable, with

$1-\dfrac{\mathrm dv}{\mathrm dx}=v^2\implies\dfrac{\mathrm dv}{1-v^2}=\mathrm dx$

On the left, we can split into partial fractions:

$\dfrac12\left(\dfrac1{1-v}+\dfrac1{1+v}\right)\mathrm dv=\mathrm dx$

Integrating both sides gives

$\dfrac{\ln|1-v|+\ln|1+v|}2=x+C$

$\dfrac12\ln|1-v^2|=x+C$

$1-v^2=e^{2x+C}$

$v=\pm\sqrt{1-Ce^{2x}}$

Now solve for $y(x)$ :

$x-y+1=\pm\sqrt{1-Ce^{2x}}$

$\boxed{y=x+1\pm\sqrt{1-Ce^{2x}}}$

3 0

3 years ago