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Vsevolod [243]
3 years ago
11

How do I solve x^2+y^2=41 y=4x-11

Mathematics
1 answer:
Paul [167]3 years ago
3 0
Download photomath on your phone and it will solve it for you and give you answer too
You might be interested in
Try it!
Aleksandr-060686 [28]

Answer:

<u><em>Answer is below</em></u>

Step-by-step explanation:

4 0
3 years ago
Which two quadrilaterals are reflections of each other across the y-axis?
ICE Princess25 [194]

Answer:

Step-by-step explanation:

When you reflect a point across the y-axis, the y-coordinate remains the same, but the x-coordinate is taken to be the additive inverse. The reflection of point (x, y) across the y-axis is (-x, y).

Hope this helped.

7 0
3 years ago
henry hit 3 more than half as many home runs as jack hit last season. if henry hit a total of 8 home runs, how mnay did jack hit
sveta [45]
Answer:
Jack hit 10 home runs.

Explanation:
Jack hit 10 home runs, Henry hit 3 more than half.

Half of 10 is 5 add 3 to reach the sum of Henry’s home runs.
5 0
3 years ago
A 64-ounce container of coconut milk costs 3.32. A 120-ounce container of the same milk costs 5.90
Marysya12 [62]

Answer: It would be cheaper to by the 120 ounce, instead of two 64 ounces.

Step-by-step explanation:

That’s because it would be $6.64, for two 64 ounces. $5.90 is cheaper than $6.64. Therefore it would be cheaper to get the 120 ounce.

5 0
3 years ago
I really need help with this question
Naily [24]

Answer:

7/12.

Step-by-step explanation:

Let's go over the details.

Box A contains 3, 5, and 6, and 9. Box B contains 1, 4, and 7.

Only 1 card is picked from each box.

For the sake of simplicity, let's take a step back. Let's assume that the actual values of the cards only indicate whether they are odd or even. So, we can replace the values with O for odd and E for even.

Box A now contains O(3), O(5), E(6), O(9), and Box B now contains O(1), E(4), O(7).

From logic, we can deduce that O + O = E, E + E = E, and E + O = O.

We're only looking at the combinations that result in an even sum. So we're specifically looking for OO and EE combinations.

Now that we know what we're looking for, we can go back to original problem.

There are a total of 12 possibilities if you use the values of the cards. Therefore, there is a \frac{1}{12} chance that you will get a specific combination.

To find the probability of getting a specific combination, you need to multiply the number of cards from Box A with the number of cards in Box B. So the possible combinations would be:

(3,1), (5,1), (6,1), (9,1), (3,4), (5,4), (6,4), (9,4), (3,7), (5,7), (6,7), (9,7)

But since we're looking only for (O,O) and (E,E) combinations, we can ignore all of the (O,E) and (E,O) combinations.

(3,1), (5,1), (9,1), (6,4), (3,7), (5,7), (9,7)

These account for 7 of the 12 possible combinations.

So the probability of getting two numbers whose sum is even, or (to put it in simpler terms) the probability of getting an OO or EE combination, is \frac{7}{12}.

4 0
3 years ago
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