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3 years ago

How many positive integers less than 250 are multiples of 5, but not multiples of 10?

Mathematics

2 answers:

SashulF [63]3 years ago

8 0

Answer:

Step-by-step explanations: sometimes 5 times (number) is 250 but in this case your teacher did not tell you to multiply it by 10 but instead multiplying it by 5

oksian1 [2.3K]3 years ago

4 0

Answer:

Step-by-step explanation:

Positive integers less than 250 are 1, 2, 3, ..., 248, 249.

Positive multiples of 5 that are not multiples of 10 are integers than end in 5.

5, 15, 25, 35, ...

In each group of 10 consecutive integers, 1 through 10, 11 through 20, etc. there is one integers ending in 5. From 1 to 250, there are 25 groups of 10 consecutive integers, so there are 25 integers that end in 5.

Answer: 25

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Write the equation in standard form. Identity the center and radius. x² + y2 + 8x-4y-7=0

lorasvet [3.4K]

Answer:

equation;

$(x + 4) {}^{2} + (y - 2) {}^{2} = 27$

Center (-4,2)

Radius is

$3 \sqrt{3}$

Step-by-step explanation:

Since the x^2 and y^2 have the same coeffiecent this will be a circle in a form of

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

Where (h,k) is center

r is the radius

So first we group like Terms together

${x}^{2} + 8x + {y}^{2} - 4y - 7 = 0$

Add 7 to both sides

${x}^{2} + 8x + {y}^{2} - 4y = 7$

$( {x}^{2} + 8x) +( {y}^{2} - 4y) = 7$

Since the orginal form of the equation of the circle has a perfect square we need to complete the square for each problem

$(\frac{8}{2} ) {}^{2} = 16$

and

$( - \frac{4}{2} ) {}^{2} = 4$

so we have

${x}^{2} + 8x + 16 + {y}^{2} - 4y + 4y = 7 + 16 + 4$

${x}^{2} + 8x + 16 + {y}^{2} - 4y + 4y = 27$

$(x + 4) {}^{2} + (y - 2) {}^{2} = 27$

To find our center, h is -4 and k is 2

so the center is (-4,2)

The radius is

$\sqrt{27} = 3 \sqrt{3}$

So the radius is 3 times sqr root of 3.

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2 years ago

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klio [65]

C I guess sorry if I get you a low score

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Mkey [24]

Answer:

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slamgirl [31]

Answer:

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Step-by-step explanation:

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