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Anastaziya [24]
3 years ago
11

Angel has $85 to spend . He spent $20 on food and saved the rest of his money for games. each game cost $5 Write an inequality t

o represent how many games , g Angel can play?
Mathematics
1 answer:
galina1969 [7]3 years ago
8 0

Answer:

85 = 5x + 20 is the inequality.

Step-by-step explanation:

Why is it right? Because I said so.

Just kidding. Here:

The cost of the games, 5, is multiplied by x, which is the amount of games that Angel will buy. To solve:

Subtract 20 from both sides:

65 = 5x

Divide 65 by 5.

The answer of the question is: Angel can buy 13 games. (Angel! That's a waste of 65 bucks!)

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One bar of candy A and two bars of candy B have 774 calories. Two bars of candy A and one bar of candy B contains 786 calories.
vampirchik [111]
◆ Define the variables:

Let the calorie content of Candy A = a
and the calorie content of Candy B = b

◆ Form the equations:

One bar of candy A and two bars of candy B have 774 calories. Thus:

a + 2b = 774

Two bars of candy A and one bar of candy B contains 786 calories

2a + b = 786

◆ Solve the equations:

From first equation,
a + 2b = 774
=> a = 774 - 2b

Put a in second equation
2×(774-2b) + b = 786
=> 2×774 - 2×2b + b = 786
=> 1548 - 4b + b = 786
=> -3b = 786 - 1548
=> -3b = -762
=> b = -762/(-3) = 254 calorie

◆ Find caloric content:

Caloric content of candy B = 254 calorie

Caloric content of candy A = a = 774 - 2b = 774 - 2×254 = 774 - 508 = 266 calorie
5 0
3 years ago
Select the three equations that pass through the points (-4, -16) and (5,2).
dezoksy [38]

Answer:

The equation of the line is y-2 = 2(x - 5), that passes through points (-4, -16)\ and\ (5,2)

Step-by-step explanation:

Given points are (-4, -16)\ and\ (5,2)

We need to write the equation of the line that passes through these points

First, we will find slope of this line

m=\frac{y_2-y_1}{x_2-x_1}\\\\m=\frac{2-(-16)}{5-(-4)}\\\\m=\frac{2+16}{5+4}\\\\m=\frac{18}{9}=2

So, our slope is 2. Now, we will plug the point (5,2) in the equation below.

(y-y_1)=m(x-x_1)\\y-2=2(x-5)\\

So, the equation of the line is y-2 = 2(x - 5).

5 0
3 years ago
Read 2 more answers
Help and explain !!!!!!!!!!!!!
Luda [366]

Answer:

6x - 15

Step-by-step explanation:

f(x) = 2x + 3 and g(x) = 3x - 9.

f(g(x))=2(3x-9)+3\\f(g(x))=6x-18+3\\f(g(x))=6x-15

7 0
3 years ago
PLzzzz HELP I will BRAINLYIST<br><br> is the relation a function (-5,2) (7,7) (3,6) (1,7)
ikadub [295]

Answer:

yes

Step-by-step explanation:

every input has exactly one output.

3 0
3 years ago
Read 2 more answers
Find the directional derivative of the function at the given point in the direction of the vector v. h(r, s, t) = ln(3r + 6s + 9
vodka [1.7K]

Answer:

D_\overrightarrow{\rm v} h(r,s,t)=(\frac{2}{7}\overrightarrow{\rm i})*(\frac{1}{r+2s+3t})+(\frac{6}{7}\overrightarrow{\rm j})*(\frac{2}{r+2s+3t})+(\frac{3}{7}\overrightarrow{\rm k})*(\frac{3}{r+2s+3t})

Step-by-step explanation:

First,  let’s check to see if the direction vector is a unit vector:

||v||=\sqrt{(14)^{2} +(42)^{2} +(21)^{2}   } =\sqrt{2401} =49

It’s not a unit vector. therefore let's divide the vector by its magnitude in order to convert it into a unit vector:

\overrightarrow{\rm v}=(\frac{14}{49}\overrightarrow{\rm i})+(\frac{42}{49}\overrightarrow{\rm j})+(\frac{21}{49}\overrightarrow{\rm k})= (\frac{2}{7}\overrightarrow{\rm i})+(\frac{6}{7}\overrightarrow{\rm j})+(\frac{3}{7}\overrightarrow{\rm k})

Now, the directional derivative is given by:

D_\overrightarrow{\rm v} h(r,s,t)=\frac{\partial h(r,s,t)}{\partial r}\overrightarrow{\rm i} + \frac{\partial h(r,s,t)}{\partial s}\overrightarrow{\rm j} + \frac{\partial h(r,s,t)}{\partial t}\overrightarrow{\rm k}

So let's calculate the partial derivates:

\frac{\partial }{\partial r} ln(3r+6s+9t)=\frac{3}{3r+6s+9t}=\frac{1}{r+2s+3t}

\frac{\partial }{\partial s} ln(3r+6s+9t)=\frac{6}{3r+6s+9t}=\frac{2}{r+2s+3t}

\frac{\partial }{\partial t} ln(3r+6s+9t)=\frac{9}{3r+6s+9t}=\frac{3}{r+2s+3t}

Therefore:

D_\overrightarrow{\rm v} h(r,s,t)=(\frac{2}{7}\overrightarrow{\rm i})*(\frac{1}{r+2s+3t})+(\frac{6}{7}\overrightarrow{\rm j})*(\frac{2}{r+2s+3t})+(\frac{3}{7}\overrightarrow{\rm k})*(\frac{3}{r+2s+3t})

3 0
3 years ago
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