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Rudik [331]
3 years ago
13

What expressions are equivalent to 3 1/4- (-5/8)

Mathematics
2 answers:
Mashcka [7]3 years ago
8 0
3 1/4 + (5/8)

That is what is equivalent to 3 1/4 -(-5/8)
I hope that helped you
Verizon [17]3 years ago
7 0

Answer:

3\frac{7}{8}

Step-by-step explanation:

The given expression is 3\frac{1}{4} - (\frac{-5}{8})

We need to simplify the fractions.

Let's convert the mixed number 3\frac{1}{4} to improper fraction.

3\frac{1}{4} = \frac{4.3 +1}{4} = \frac{12 + 1}{4} = \frac{13}{4}

So, \frac{13}{4} -(\frac{-5}{8} )

= \frac{13}{4} +(\frac{5}{8} )

Now we have to find the least common divisor (LCD) of 4 and 8. So that we can simplify the fractions.

The LCD of 4 and 8 is 8.

So make the denominators as 8.

= \frac{13(2)}{4(2)} + \frac{5}{8}

= \frac{26}{8} + \frac{5}{8}

Now the denominators became the same. Now we can simplify the numerator and keep the denominator as it is.

=  \frac{26 + 5}{8}

=  \frac{31}{8}

Now we can convert this improper fraction to mixed number

= 3\frac{7}{8}

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Given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle
Anastasy [175]

Answer:

Part 1) False

Part 2) False

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

(x-h)^{2} +(y-k)^{2}=r^{2}

where

(h,k) is the center and r is the radius

In this problem the distance between the center and a point on the circle is equal to the radius

The formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

Part 1) given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

(x+3)^{2} +(y-4)^{2}=r^{2}

Find the distance (radius) between the center (-3,4) and (-6,2)

substitute in the formula of distance

r=\sqrt{(2-4)^{2}+(-6+3)^{2}}

r=\sqrt{(-2)^{2}+(-3)^{2}}

r=\sqrt{13}\ units

The equation of the circle is equal to

(x+3)^{2} +(y-4)^{2}=(\sqrt{13}){2}

(x+3)^{2} +(y-4)^{2}=13

Verify if the point (10,4) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=10,y=4

substitute

(10+3)^{2} +(4-4)^{2}=13

(13)^{2} +(0)^{2}=13

169=13 -----> is not true

therefore

The point is not on the circle

The statement is false

Part 2) given the center of the circle (1,3) and a point on the circle (2,6), (11,5) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

(x-1)^{2} +(y-3)^{2}=r^{2}

Find the distance (radius) between the center (1,3) and (2,6)

substitute in the formula of distance

r=\sqrt{(6-3)^{2}+(2-1)^{2}}

r=\sqrt{(3)^{2}+(1)^{2}}

r=\sqrt{10}\ units

The equation of the circle is equal to

(x-1)^{2} +(y-3)^{2}=(\sqrt{10}){2}

(x-1)^{2} +(y-3)^{2}=10

Verify if the point (11,5) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=11,y=5

substitute

(11-1)^{2} +(5-3)^{2}=10

(10)^{2} +(2)^{2}=10

104=10 -----> is not true

therefore

The point is not on the circle

The statement is false

7 0
3 years ago
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