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Illusion [34]
3 years ago
6

Help me please!!!

Mathematics
1 answer:
Jlenok [28]3 years ago
3 0

Answer:

10

Step-by-step explanation:

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The diagonals of quadrilateral EFGH intersect at D(?3,4). EFGH has vertices at E(3,7) and F(?4,5). What must be the coordinates
Katen [24]

Answer:

  G(3, 1), H(2, 3)

Step-by-step explanation:

When D is the midpoint of EG, it means ...

  D = (E + G)/2

or

  G = 2D -E = 2(3,4) -(3,7) = (2·3-3, 2·4-7) = (3, 1)

Likewise, H is ...

  H = 2D -F = 2(3,4) -(4,5) = (2·3-4, 2·4-5) = (2, 3)

5 0
3 years ago
An organization has members who possess IQs in the top 4% of the population. If IQs are normally distributed, with a mean of 100
murzikaleks [220]

Answer:

The minimum IQ score will be "126".

Step-by-step explanation:

The given values are:

Mean

\mu = 100

Standard deviation

\sigma=15

Now,

⇒ P(z>x)=4 \ percent \ i.e., 0.04

⇒ P(z>\frac{x- \mu}{\sigma} )=0.04

⇒ 1-P(z \leq \frac{x-100}{15})=0.04

⇒ \frac{x-100}{15}=z0.96

            =NORMDIS(z=0.96)

            =1.751

⇒ x=100+15\times 1.751

      =126.265 \ i.e., 126

7 0
3 years ago
If 5 bulbs are $24.95 how much money is 1 bulb?
Elden [556K]

Answer:

4.99

Step-by-step explanation:

If you divide:

$24.95/5bulbs

you get

4.99/1 bulb  

5 0
3 years ago
Marshall divides his money into four categories. He saves 1/3 of his money. He gives 1/6 of his money to a charity. He uses 1/14
Len [333]
1/3 + 3/7
save + fun + extra

(7/21 + 9/21) = 16/21 on save and fun
7 0
3 years ago
Read 2 more answers
What are the possible numbers of positive, negative, and complex zeros of f(x) = −3x4 −
Anna007 [38]

Answer:

b.

Step-by-step explanation:

We have to look at sign changes in f(x) to determine the possible positive real roots.

f(x)=-3x^4-5x^3-x^2-8x+4

There is only one sign change here, between the -8x and the +4.  So that means there is only 1 possible real positive root.

Now we have to look at sign changes in f(-x) to determine the possible negative real roots.

f(-x)=-3x^4+5x^3-x^2+8x+4

There are 3 sign changes here.  That means there are either 3 negative roots or 3-2 = 1 negative root.  So we have:

1 positive

3 or 1 negative

We need to pair them up now with all the possible combinations.

If we have 1 positive and 1 negative, we have to have 2 imaginary

If we have 1 positive and 3 negative, we have to have 0 imaginary

Keep in mind that the total number or roots--positive, negative, imaginary--have to add up to equal the degree of the polynomial.  This is a 4th degree polynomial, so we will have 4 roots.

7 0
3 years ago
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