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2 years ago

What is the value of the expression below when y = 3? 10y - 8

Mathematics

1 answer:

ycow [4]2 years ago

3 0

Answer:

Step-by-step explanation:

<h3>Given, y = 3 </h3>

substitute this value in given expression

=> 10y - 8

=> 10(3) - 8

=> 30 - 8

=> 22...ans

<h2>HOPE IT HELPS U!!!!</h2>

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HELP ME WITH 9, 11, 12

mart [117]

Answer:

Question 9

Name the ordered pairs in the table: 3, 4, 5, 6, 7

Ordered pairs: (1,3), (2,4), (3,5), (4,6), (5,7)

Graph the equation: the attachment

Step-by-step explanation:

I don't know how to do questions 11 and 12.

Also sorry if these aren't correct.

8 0

2 years ago

Find the solution set of the inequality

Brilliant_brown [7]

Answer:

$\boxed{\textsf{ Hence the solution set is$ \bf{\bigg( \dfrac{33}{4},\infty \bigg) }$}}.$

Step-by-step explanation:

A inequality is given to us and we need to find the solution set. So the given inequality to us is ,

-4x + 35 >2

$\bf\implies -4x + 35 > 2 \\\\\bf\implies -4x > 2-35 \\\\\bf \implies -4x > -33 \\\\\bf \implies x >\dfrac{-33}{-4}\\\\\bf \implies x > \dfrac{33}{4} \\\\\bf\implies \boxed{\pink{\bf x \in \bigg(\dfrac{33}{4}, \infty \bigg) }}$

<h3>★ Hence the solution set is x € ( 33/4 , ∞ ).</h3>

3 0

2 years ago

A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min

sesenic [268]

$A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})$
$\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$
$A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4$

We're given that $A(0)=30$ . Multiply both sides by the integrating factor $e^{t/100}$ , then

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C$
$A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}$

Given that $A(0)=30$ , we have

$30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02$

so the amount of salt in the tank at time $t$ is

$A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}$

3 0

3 years ago

Fill in the blanks people!

MakcuM [25]

I can’t understand what you are asking for

8 0

3 years ago

(40 POINTS) PLZ HELP ASAP! DOMAIN AND RANGE! CLICK LINK

MAVERICK [17]

For finding the values, we look at the x - value given. Then we move to where it is on the graph and find its y value.

In part A, when x = -4, the y value looks to be between -3 and -4. Let's put it in the middle and estimate it at -3.5

In part B, when x = 1, the process is similar. Go to x = 1, then go to the graph, then go to the y value. That looks to be at y = -3.

In part C, when x = 4, there are three values that work. We are actually answering part D at this point because we can tell this graph is NOT a function of x. When you have an x -value that goes into the function, you have to get EXACTLY ONE thing that comes out. I bolded "EXACTLY ONE" because those words make the definition work. Two or more, not a function. One, it's a function. None, and it's not in the domain at all. There are three values of y that work, and they are - from top down, 3.5, 0, -2.5.

We answered Part D when observing part C. It's NOT a function because there is not EXACTLY ONE value of y for an x.

And finally, because it's not a function - finding the domain and range is a waste of time. You can't find the domain of something that's not a function - you need a function to have a domain.

Hope that helps.

7 0

3 years ago