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3 years ago

What is the following coordinates of (2,0) and (0,4)

Mathematics

1 answer:

34kurt3 years ago

8 0

Answer:

(2, 0) and (0, 4) are coordinates in the form of (x, y)

(2, 0) -> x = 2, y = 0

(0, 4) -> x = 0, y = 4

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Find the midpoint of CD. C= (7,-4) and D= (-8,-5)

ANTONII [103]

Answer:

( $-\frac{1}{2}$ , $-\frac{9}{2}$ )

Step-by-step explanation:

The equation of a midpoint is as follows:

$(\frac{x_{1} +x_{2}}{2} ,\frac{y_{1} +y_{2}}{2})$

The x-coordinate:

x1 = 7, x2 = -8

7+(-8) = -1

Divide by 2 to get the x-midpoint

-1/2

The y-coordinate

y1 = -4, y2 = -5

-4 + (-5) = -9

Divide by 2 to get the y-midpoint

-9/2

You get:

( $-\frac{1}{2}$ , $-\frac{9}{2}$ )

5 0

3 years ago

Read 2 more answers

jordan has 11 feet of lumber to make wooden signs. If each sign requires 2/3 of a foot of lumber, how many signs can he make?

Firdavs [7]

Answer:16 signs

Step-by-step explanation:

you have 11 feet of wood and for each sign you need 2/3 so you would do 11 divided by 2/3 and yu get 16.5 but you can't have 16.5 signs and you don't have enough for 17 signs so it would be 16 signs

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4 years ago

A student answers 80% of the questions on the math exam correctly, if she answers 40 questions correctly. How many questions are

mafiozo [28]

Hope this helps! ask if you need clarification.

6 0

4 years ago

3 X square + 3 X - 36 by 6 x square - 24 x - 30

iVinArrow [24]

Step-by-step explanation:

The solution is in the above photo

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4 years ago

Let n be an odd integer with exactly 11 positive divisors. Find the number of positive divisors of 8n^3

Tom [10]

Answer: 124

Step-by-step explanation:

As we know, the total number of positive divisors of n is the product of one more than each exponent in the prime factorization of n.

We know this product is equal to 11.

Since 11 is prime, so there can be only one number in this product, so the prime factorization of n must be $n=p^{10}$ , where p is a prime number.

We take a cube of the prime factorization of n to give us a prime factorization for $n^3$

$n^{3}=\left(p^{10}\right)^{3}=p^{30} .$

We see that $n^{3}\text{ has }$30+1=31$$ positive divisors.

Now, $8n^3=2^3\times p^{30}$

Totald divisiors = $(3+1)(30+1)=4\times31=124$

Hence, the number of positive divisors of $8n^3=124$

4 0

3 years ago