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vichka [17]
3 years ago
14

Tom’s telephone company charges a flat fee of $33.00 a month, plus $0.10 a minute for all long-distance calls made during the mo

nth. Tom’s bill for the month of June was $53.00 before taxes. Which equation can be used to determine how many long-distance minutes, x, Tom used during the month of June?
Mathematics
1 answer:
OleMash [197]3 years ago
6 0
F(x)= 33+.10x. Since its 10 cents per minute, x would represent the number of minutes Tom uses.
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5 0
3 years ago
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Find the measure of the number two angles in each figure. Justify each answer​
BigorU [14]
11:
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8 0
3 years ago
Sally and Karl work at different jobs. Sally earns $7 per hour and Karl earns $5 per hour. They each earn the same amount per we
Lisa [10]

Answer:

Karl works 7 hours a week

Step-by-step explanation:

Step 1: Determine total amount that Sally earns

Total amount Sally earns=rate per hour×number of hours worked(h)

where;

Rate per hour=$7 per hour

Number of hours worked=h

Replacing;

Total amount Sally earns=(7×h)=7 h

Step 2: Determine total amount Karl earns

Total amount Karl earns=rate per hour×number of hours worked

where;

rate per hour=$5

number of hours worked=2 more than Sally=h+2

replacing;

Total amount Karl earns=5(h+2)

Step 3: Equate Sally's total earnings to Karl's total earnings and solve for h

7 h=5(h+2)

7 h=5 h+10

7 h-5 h=10

2 h=10

h=10/2

h=5

Karl works (h+2) hours=(5+2)= 7 hours

Karl works 7 hours a week

7 0
4 years ago
PLEASE HELP!! I do not understand this. And can you explain it for me when I use it again?
damaskus [11]
Let's start b writing down coordinates of all points:
A(0,0,0)
B(0,5,0)
C(3,5,0)
D(3,0,0)
E(3,0,4)
F(0,0,4)
G(0,5,4)
H(3,5,4)

a.) When we reflect over xz plane x and z coordinates stay same, y coordinate  changes to same numerical value but opposite sign. Moving front-back is moving over x-axis, moving left-right is moving over y-axis, moving up-down is moving over z-axis.

A(0,0,0)
Reflecting
A(0,0,0)

B(0,5,0)
Reflecting
B(0,-5,0)

C(3,5,0)
Reflecting
C(3,-5,0)

D(3,0,0)
Reflecting
D(3,0,0)


b.)
A(0,0,0)
Moving
A(-2,-3,1)

B(0,-5,0)
Moving
B(-2,-8,1)

C(3,-5,0)
Moving
C(1,-8,1)

D(3,0,0)
Moving
D(1,-3,1)
7 0
3 years ago
I need help with my algebra 2. I also want to know how to solve this.
Lisa [10]

Add 1 to both sides:

\sqrt{x+3} = x+1

In cases like this, we have to remember that a root is always positive, so we can square both sides only assuming that

x+1 \geq 0 \iff x \geq -1

Under this assumption, we square both sides and we have

x+3 = (x+1)^2 \iff x+3 = x^2+2x+1 \iff x^2+x-2 = 0

The solutions to this equation are

x = -2,\ x=1

But since we can only accept solutions greater than -1, we discard x=-2 and accept x=1.

In fact, we have

x=-2 \implies \sqrt{-2+3}-1=0\neq -2

and

x=1 \implies \sqrt{4}-1=1

which is the only solution.

4 0
4 years ago
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