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Marina86 [1]
2 years ago
11

Anybody know the real answer to this?

Mathematics
1 answer:
maks197457 [2]2 years ago
8 0

Look at the picture.

Use the Pythagorean theorem:

x^2=(3+2)^2+(4+5)^2\\\\x^2=5^2+9^2\\\\x^2=25+81\\\\x^2=106\to x=\sqrt{106}\\\\x\approx10.3

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Find the intersection points of the linear and quadratic functions shown below f(x)=2x-5 g(x)= x2+2x-21
Mice21 [21]

Answer:

(-4,-13) and (4,3) the intersection points.

Step-by-step explanation:

Intersection point of two functions is a common point which satisfies both the functions.

Given functions are,

f(x)=2x-5

g(x)=x^2+2x-21

For a common point of these functions,

f(x)=g(x)

2x-5=x^2+2x-21

-5=x^2-21

0=x^2-16

x^2=16

x=-4,4

For x=-4,

f(-4)=g(-4)=2(-4)-5

                       =-13

For x=4,

f(4)=g(4)=2(4)-5

                  =3

Therefore, (-4,-13) and (4,3) the intersection points.

3 0
2 years ago
Simple interst for $750 at 6.5 interest for 3 years
ser-zykov [4K]
Considering that you mean 6.5% interest yearly, the interest gained would be:
$146.25

and the total amount would be $896.25

Hope this helped!
4 0
2 years ago
Read 2 more answers
A group of 123 students went on a field trip to
irakobra [83]

Answer:

8.2

Step-by-step explanation:

If each student in the group collected 15 shells, we know that there are 123 students on the field trip, so we can take 123 and divide it by 15 to get 8.2 or 41/5.

7 0
3 years ago
Toby's piggy bank contains only 5c and 10c coins. If it contains 65 coins with a total value of 3.80, find the number of each ty
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The positive is 2 hknshue dska
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3 years ago
If the product of two positive fractions a and b is 15/56, find three pairs of possible values for a and b
Andrews [41]

Answer:

First possible pairs = (3/7)*(5/8) = 15/56

second possible pairs = (3/4)*(5/14) = 15/56

third possible pairs = (1/2)*(15/28) = 15/56

Step-by-step explanation:

According to the question,

a x b = 15/56

three possible pairs will be as follows:

Before proceeding to make possible pairs, we have to know the prime factorization of the numbers -

15 = 1, 3, 5, 15

56 = 1, 2, 4, 7, 8, 14, 28, 56

As we need three possible pairs, we have to check which pairs take us to 15/56. Again, since the fraction is positive, therefore, the fractions will be proper. Therefore, 3/2, or 5/4 will not be counted. Therefore,

First possible pairs = (3/7)*(5/8) = 15/56

second possible pairs = (3/4)*(5/14) = 15/56

third possible pairs = (1/2)*(15/28) = 15/56

fourth possible pairs = (3/8)*(5/7) = 15/56

So, we can get those four possible pairs. Among those, first 3 pairs are different. Therefore, those are possible pairs.

4 0
2 years ago
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