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3 years ago

5

I need to know what the percent of decrease is?

2 answers:

lys-0071 [83]3 years ago

6 0

The percent of decrease from 20 to 16 is 20%

PolarNik [594]3 years ago

4 0

B 20% accidentally put that in comments

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64 ounces in 8 cups unit rate

Veseljchak [2.6K]

So the unit rate is basically how many ounces in 1 cup?

So since there are 64 ounces in 8 cups, you can just divide by 8, to find how many sets of 8 there are in 64.

64 = 8

64 / 8 = 8 / 8

8 = 1

So there are 8 ounces in 1 cup, which is your unit rate.

7 0

3 years ago

1.) An American tourist changed US $925 into Trinidad and Tobago currency at the exchange rate US $1.00= TT $ 4.25

storchak [24]

Step-by-step explanation:

1)a) 925×4.25=3931,25

b) 1806.25:4.25=425

I don't know second one.

3)a 55×1.12=61.6

b)

8 0

3 years ago

Which number line shows the solution set for the inequality -4x+7>15

Ira Lisetskai [31]

First, let's simplify the inequality as much as we can:

-4x + 7 > 15
-4x > 8

Divide by -4:
x < -2

3 0

3 years ago

Isaiah ran 100 meters on a 200 meter track. He started running on the 19 meter mark. On what Mark did he finish his run

guapka [62]

Answer:

119

Step-by-step explanation:

19+100=119

5 0

3 years ago

Read 2 more answers

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

2 years ago