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Bond [772]
2 years ago
9

Can someone help me please?

Mathematics
2 answers:
Damm [24]2 years ago
8 0
For the first question you shade in the bar to the 20% mark as she has saved 20% of the money needed for her jacket. To complete the bar you add the amount of how much each percentage is in dollars. We know 20%=$15 and the percentage goes up 20% each time so you keep adding $15 on. For question 2 the bar is half way between 0 and 20 and half of 20 is 10. As 10 is half of 20 we half 15 to get $7.50
Alekssandra [29.7K]2 years ago
4 0

Answer:

i dont know

Step-by-step explanation:

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The perimeter of a square is 20 ft. If you increase the length of the square by 2 feet and decrease the width by 1 foot, what is
Mariulka [41]
Technically, you can't do that because a square has equal side lengths
4 0
3 years ago
Read 2 more answers
Celia and Ryan are starting a nutrition program.Celia currently consumes 1200 calories a day and will increase that number by 10
BartSMP [9]

Answer:

Celia consumes 100

Step-by-step explanation:

Because thats just right

3 0
3 years ago
How many different ordered pairs satisfy both x^{2} + y^{2} = 100 and x^{2} + 2y^{2} = 108?
Dominik [7]
\begin{cases}x^2+y^2=100\\x^2+2y^2=108\end{cases}\\\\\\
\begin{cases}x^2+y^2=100\\x^2+y^2+y^2=108\end{cases}\\\\\\100+y^2=108\\\\y^2=108-100\\\\y^2=8\qquad|\sqrt{(\ldots)}\\\\y=-\sqrt{8}\qquad\vee\qquad y=\sqrt{8}\\\\\boxed{y=-2\sqrt{2}\qquad\vee\qquad y=2\sqrt{2}}

We know that y^2=8 so:

x^2+y^2=100\\\\x^2+8=100\\\\x^2=100-8\\\\x^2=92\qquad|\sqrt{(\ldots)}\\\\
x=-\sqrt{92}\qquad\vee\qquad x=\sqrt{92}\\\\x=-\sqrt{4\cdot23}\qquad\vee\qquad x=\sqrt{4\cdot23}\\\\\boxed{x=-2\sqrt{23}\qquad\vee\qquad x=2\sqrt{23}}

As we see there are 4 such pairs:

x=-2\sqrt{23}\qquad y=-2\sqrt{2}\\\\x=2\sqrt{23}\qquad y=-2\sqrt{2}\\\\
x=-2\sqrt{23}\qquad y=2\sqrt{2}\\\\x=2\sqrt{23}\qquad y=2\sqrt{2}


5 0
3 years ago
The answer is 12/70 you can simplify it
4vir4ik [10]

Answer:

6/35

Step-by-step explanation:

12 divided by 2 is 6 and 70 divided by 2 is 35

4 0
1 year ago
Please help me with this question???
Serjik [45]
The correct answer is the third option.
5 0
3 years ago
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