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Sav [38]
3 years ago
5

URGENT!

Mathematics
1 answer:
Alina [70]3 years ago
4 0
4.125 x 14 = 57.75 So 57.75 is the answer and I have to show my work
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Replace ∗ with a monomial so that the derived equality will be an identity: (15y + ∗)2 = 225y2+12x3y+0.16x6
kobusy [5.1K]

Answer:

<u>∗ = 0.4x³</u>

Step-by-step explanation:

(15y + ∗)² = 225y²+12x³y+0.16x⁶

<u>Note:</u>

225y² = 15y * 15y = (15y)²

12x³y = 2 * 15y * 0.4x³

0.16x⁶ = 0.4x³ * 0.4x³ = (0.4x³)²

So, by factoring the right hand side:

225y²+12x³y+0.16x⁶ = (15y + 0.4x³)²

By comparing the left hand side with (15y + 0.4x³)²

<u>So, ∗ should be replaced with the monomial 0.4x³</u>

6 0
3 years ago
Square root of 81 minus square root of negative 48 answer in a + bi form
galina1969 [7]

Answer:

9-6.92820323i Nothing else can be done.

Step-by-step explanation:

-48 is not a perfect square but 81 is a square. When you try to square -48 it comes to be 6.92820323i.

4 0
3 years ago
Teps to go<br> h(x) = x2 + 5x + 6<br> write in factored form
Ronch [10]

Answer:

h(x) = (x + 2)(x + 3)

Done .....

♥️♥️♥️

4 0
3 years ago
Someone help me figure this out
Lena [83]

-6, because you can see a pattern in which the x coordinate of each pair is reversed to the opposite value, from positive to negative.

5 0
3 years ago
Read 2 more answers
Suppose that x has a binomial distribution with n = 201 and p = 0.45. (Round np and n(1-p) answers to 2 decimal places. Round yo
ycow [4]

Answer:

a) It can be used because np and n(1-p) are both greater than 5.

Step-by-step explanation:

Binomial distribution and approximation to the normal:

The binomial distribution has two parameters:

n, which is the number of trials.

p, which is the probability of a success on a single trial.

If np and n(1-p) are both greater than 5, the normal approximation to the binomial can appropriately be used.

In this question:

n = 201, p = 0.45

So, lets verify the conditions:

np = 201*0.45 = 90.45 > 5

n(1-p) = 201*(1-0.45) = 201*0.55 = 110.55 > 5

Since both np and n(1-p) are greater than 5, the approximation can be used.

3 0
3 years ago
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