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Tamiku [17]
3 years ago
14

Please Help Evaluate 5c - 3d + 11 when c=7 and d=8

Mathematics
2 answers:
musickatia [10]3 years ago
4 0
<h3>Hey there! </h3><h3>If, c = 7 and d = 8</h3><h3>Then, let's solve the equation, shall we? </h3><h3>5(7) -3(8) + 11</h3><h3>5(7) = 35</h3><h3>35 - 3(8) + 11 </h3><h3>3(8) = 24</h3><h3>35 - 24 + 11</h3><h3>35 - 24 = 11 </h3><h3>11 + 11 = = 22</h3><h3>Answer: 22</h3><h3>Good luck on your assignment and enjoy your day! </h3><h3>~LoveYourselfFirst:)</h3>
allochka39001 [22]3 years ago
3 0
The answer would be 22.

You would use the order of operations to solve this problem,

First you do the parenthesis, there are none so you skip this

Then the exponents, none so skip it

Then you do multiplication and division from left to right

5•7 is 35, 3•8 is 24

So the problem is now
35-24+11

After multiplication comes addition and subtraction from left to right

35-24 is 11
11+11 = 22

So the answer to (5x7)-(3x8)+11 is 22
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pentagon [3]

Answer:

3.33

4.33

TVE figures is that

6 0
3 years ago
How many liters of water must be added to 15 liters of 40 % sugar syrup to obtain 30 % sugar syrup?
Ivanshal [37]

Answer:

The amount of water need to be added is 5 liters.

Step-by-step explanation:

Let's "x" be amount of water in (liters) added to 15 liters of 40% of sugar syrup.

Now find the amount of sugar syrup = 40% of 15

= 0.4 × 15

The amount of sugar syrup = 6 Liters

To dilute  30% we need to find amount of water to be added.

So,

30% of (15 + x) = 6

0.3 × (15 + x) = 6

4.5 + 0.3x = 6

0.3x = 6 - 4.5

0.3x = 1.5

Dividing both sides, by 0.3, we get

x = 5

So, the amount of water need to be added is 5 liters.

6 0
3 years ago
Read 2 more answers
Which inequality is true? 1.5 greater-than 2 and one-half
Travka [436]

The inequality negative 3 and one-half greater-than negative 4.5 or -3.5 > -4.5 is true.

<h3>What is inequality?</h3>

It is defined as the expression in mathematics in which both sides are not equal they have mathematical signs either less than or greater than known as inequality.

We have inequalities given:

1.5 > 2\dfrac{1}{2}

or

1.5 > 2.5  (false)

1/2 > 0.5

0.5 > 0.5 (false)

-2.5 > -1.5 (false)

-3 \dfrac{1}{2} > -4.5

-3.5 > -4.5  (true)

Thus, the inequality negative 3 and one-half greater-than negative 4.5 or -3.5 > -4.5 is true.

Learn more about the inequality here:

brainly.com/question/19491153

#SPJ1

7 0
1 year ago
Expression that represent the product of q and 5 divided by 2
Anni [7]
The answer is
5q/2 or q×5/2
8 0
3 years ago
Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite
zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt  be a gram/L

Let the amount salt in the tank at any time t be Q(t).

\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}

Incoming rate = (a g/L)×(1 L/min)

                       =a g/min

The concentration of salt in the tank at any time t is = \frac{Q(t)}{10}  g/L

Outgoing rate = (\frac{Q(t)}{10} g/L)(1 L/ min) \frac{Q(t)}{10} g/min

\frac{dQ}{dt} = a- \frac{Q(t)}{10}

\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt

Integrating both sides

\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt

\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c        [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c

\Rightarrow -log|10a-15|-2=c

Therefore ,

-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2 .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0  in equation (1) we get

- log|10a|= -log|10a-15| -2

\Rightarrow- log|10a|+log|10a-15|= -2

\Rightarrow log|\frac{10a-15}{10a}|= -2

\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}

\Rightarrow 1-\frac{15}{10a} =e^{-2}

\Rightarrow \frac{15}{10a} =1-e^{-2}

\Rightarrow \frac{3}{2a} =1-e^{-2}

\Rightarrow2a= \frac{3}{1-e^{-2}}

\Rightarrow a = 1.73

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0
3 years ago
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