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3 years ago

14

Kim drew a map of the Mississippi River. The scale was 3 cm represents 120 miles. The actual length of the Mississippi River is

2320 miles. What is the the length of the Mississippi River on the map?

1 answer:

Citrus2011 [14]3 years ago

7 0

Answer:

58cm

Step-by-step explanation:

2320/120

19.33...*3

58cm

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Rachael bought a snorkel and ear plugs for shallow diving. The snorkel cost \$27.69 , and the ear plugs cost \$2.19 . The sales

JulijaS [17]

Answer:

The answer is 10.12

Step-by-step explanation:

You add both numbers together and then subtract the sum to the 40 dollars

7 0

3 years ago

The total source voltage in the circuit is 6-3i V. What is the voltage at the middle source

Bezzdna [24]

Answer:

<em>The voltage at the middle source is</em> $(2-4\mathbf{i})\ V$

Step-by-step explanation:

<u>Voltage Sources in Series</u>

When two or more voltage sources are connected in series, the total voltage is the sum of the individual voltages of each source.

The figure shown has three voltage sources of values:

$2 + 6\mathbf{i}$

$a + b\mathbf{i}$

$2 - 5\mathbf{i}$

The sum of these voltages is:

$V_t=4+a+(6+b-5)\mathbf{i}$

Operating:

$V_t=4+a+(1+b)\mathbf{i}$

We know the total voltage is $6-3\mathbf{i}$ , thus:

$4+a+(1+b)\mathbf{i}=6-3\mathbf{i}$

Equating the real parts and the imaginary parts independently:

4+a=6

1+b=-3

Solving each equation:

a = 2

b = -4

The voltage at the middle source is $(2-4\mathbf{i})\ V$

5 0

3 years ago

Find the value of y for the following system of equations 5x + 3y = 10 x = y - 6

viktelen [127]

Answer:

x = -1 , y = 5

Step-by-step explanation:

Solve the following system:

{5 x + 3 y = 10 | (equation 1)

x = y - 6 | (equation 2)

Express the system in standard form:

{5 x + 3 y = 10 | (equation 1)

x - y = -6 | (equation 2)

Subtract 1/5 × (equation 1) from equation 2:

{5 x + 3 y = 10 | (equation 1)

0 x - (8 y)/5 = -8 | (equation 2)

Multiply equation 2 by -5/8:

{5 x + 3 y = 10 | (equation 1)

0 x+y = 5 | (equation 2)

Subtract 3 × (equation 2) from equation 1:

{5 x+0 y = -5 | (equation 1)

0 x+y = 5 | (equation 2)

Divide equation 1 by 5:

{x+0 y = -1 | (equation 1)

0 x+y = 5 | (equation 2)

Collect results:

Answer: {x = -1 , y = 5

5 0

4 years ago

Identify the domain and range of the following relation. 3.7

andrey2020 [161]

<span>Answer: {(3,7),(3,8),(3,-2),(3,4),(3,1)} A.Domain:{3} Range:{-2,1,4,7,8}**** B.Domain:{-2,1,4,7,8}</span>

7 0

4 years ago

Let w(s,t)=f(u(s,t),v(s,t)) where u(1,0)=−6,∂u∂s(1,0)=5,∂u∂1(1,0)=7 v(1,0)=−8,∂v∂s(1,0)=−8,∂v∂t(1,0)=6 ∂f∂u(−6,−8)=−1,∂f∂v(−6,−8

Blababa [14]

$w(s,t)=f(u(s,t),v(s,t))$

From the given set of conditions, it's likely that you are asked to find the values of $\dfrac{\partial w}{\partial s}$ and $\dfrac{\partial w}{\partial t}$ at the point $(s,t)=(1,0)$ .

By the chain rule, the partial derivative with respect to $s$ is

$\dfrac{\partial w}{\partial s}=\dfrac{\partial f}{\partial u}\dfrac{\partial u}{\partial s}+\dfrac{\partial f}{\partial v}\dfrac{\partial v}{\partial s}$

and so at the point $(1,0)$ , we have

$\dfrac{\partial w}{\partial s}\bigg|_{(s,t)=(1,0)}=\dfrac{\partial f}{\partial 
u}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial u}{\partial s}\bigg|_{(s,t)=(1,0)}+\dfrac{\partial f}{\partial 
v}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial v}{\partial s}\bigg|_{(s,t)=(1,0)}$
$\dfrac{\partial w}{\partial s}\bigg|_{(s,t)=(1,0)}=(-1)(5)+(2)(-8)=-21$

Similarly, the partial derivative with respect to $t$ would be found via

$\dfrac{\partial w}{\partial t}\bigg|_{(s,t)=(1,0)}=\dfrac{\partial f}{\partial 
u}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial u}{\partial t}\bigg|_{(s,t)=(1,0)}+\dfrac{\partial f}{\partial 
v}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial v}{\partial t}\bigg|_{(s,t)=(1,0)}$
$\dfrac{\partial w}{\partial t}\bigg|_{(s,t)=(1,0)}=(-1)(7)+(2)(6)=5$

6 0

4 years ago