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3 years ago

HELP ASAP

Mathematics

1 answer:

lora16 [44]3 years ago

3 0

I solved it on a piece of paper. I can not answer why I got the same answer, but I hope my method helped

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Solve u^2 = -4, where u is a real number. Simplify your answer as much as possible.

alexdok [17]

Answer:

2i and -2i

Step-by-step explanation:

I'll change u to x to be easier to understand.

x^2 = -4

sqrt(x^2) = sqrt(-4)

sqrt(-4) = 2i, or -2i

8 0

2 years ago

one bucket of gravel has a mass of 7.05 kilograms. what is the mass for 20 buckets of gravel in kilograms

Westkost [7]

Answer:

tell me if am wrong.

Step-by-step explanation:

1 bucket=7.05kg

20 bucket=20*7.05=141 kg

the mass of 20 bucket is 141 kilogram(kg)

5 0

3 years ago

A square television screen has sides that are 23 inches. Find the approximate length of its diagonal to the nearest tenth of an

Lana71 [14]

Answer:

Step-by-step explanation:

6 0

4 years ago

Help please!!! I dont understand these questions<br><br><br>currently attaching photos dont delete

Katyanochek1 [597]

Answer:

b/a
16a²b²
n¹⁰/(16m⁶)
y⁸/x¹⁰
m⁷n³n/m

Step-by-step explanation:

These problems make use of three rules of exponents:

$a^ba^c=a^{b+c}\\\\(a^b)^c=a^{bc}\\\\a^{-b}=\dfrac{1}{a^b} \quad\text{or} \quad a^b=\dfrac{1}{a^{-b}}$

In general, you can work the problem by using these rules to compute the exponents of each of the variables (or constants), then arrange the expression so all exponents are positive. (The last problem is slightly different.)

1. There are no "a" variables in the numerator, and the denominator "a" has a positive exponent (1), so we can leave it alone. The exponent of "b" is the difference of numerator and denominator exponents, according to the above rules.

$\dfrac{b^{-2}}{ab^{-3}}=\dfrac{b^{-2-(-3)}}{a}=\dfrac{b}{a}$

2. 1 to any power is still 1. The outer exponent can be "distributed" to each of the terms inside parentheses, then exponents can be made positive by shifting from denominator to numerator.

$\left(\dfrac{1}{4ab}\right)^{-2}=\dfrac{1}{4^{-2}a^{-2}b^{-2}}=16a^2b^2$

3. One way to work this one is to simplify the inside of the parentheses before applying the outside exponent.

$\left(\dfrac{4mn}{m^{-2}n^6}\right)^{-2}=\left(4m^{1-(-2)}n^{1-6}}\right)^{-2}=\left(4m^3n^{-5}}\right)^{-2}\\\\=4^{-2}m^{-6}n^{10}=\dfrac{n^{10}}{16m^6}$

4. This works the same way the previous problem does.

$\left(\dfrac{x^{-4}y}{x^{-9}y^5}\right)^{-2}=\left(x^{-4-(-9)}y^{1-5}\right)^{-2}=\left(x^{5}y^{-4}\right)^{-2}\\\\=x^{-10}y^{8}=\dfrac{y^8}{x^{10}}$

5. In this problem, you're only asked to eliminate the one negative exponent. That is done by moving the factor to the numerator, changing the sign of the exponent.

$\dfrac{m^7n^3}{mn^{-1}}=\dfrac{m^7n^3n}{m}$

3 0

3 years ago

n parallelogram LMNO, MP = 21 m, LP = (y + 3) m, NP = (3y – 1) m, and OP = (2x – 1) m. What are the values of x and y? x = 10, y

seropon [69]

Answer:

The value of x = 11 and y = 2

Step-by-step explanation:

Given : parallelogram LMNO, MP = 21 m, LP = (y + 3) m, NP = (3y – 1) m, and OP = (2x – 1) m.

We have to find values of x and y.

Let P be the point of intersection of diagonals OM and LN.

In a parallelogram diagonal bisects at right angles and point of intersection divide diagonal in equal parts.

Thus, OP = MP and LP = PN

OP = MP , substitute the values, we get,

(2x-1) = 21

⇒ 2x = 22

⇒ x = 11

LP = PN , substitute the values, we get,

y + 3 = 3y -1

⇒ 3y - y = 4

⇒ 2y = 4

⇒ y = 2

Thus, the value of x = 11 and y = 2

8 0

4 years ago