All categories

2 years ago

The pic is 6 red heart and 4 green hearts

Mathematics

2 answers:

wolverine [178]2 years ago

7 0

The answer is C)
have a nice day

Basile [38]2 years ago

6 0

It's C) 0.4

Just took the test

You might be interested in

Combine the like terms to simplify the expression: 9+3+17xy(2)+8y(3)+10x –13xy(2) – 9x – 6y(3)

seraphim [82]

9 + 3 + 17xy² + 8y³ + 10x - 13xy² - 9x - 6y³
12 + 17xy² + 8y³ + 10x - 13xy² - 9x - 6y³
17xy² - 13xy² + 10x - 9x + 8y³ - 6y³ + 12
4xy² + x + 2y³ + 12

The answer is A.

7 0

2 years ago

For (x - 1)(x - 7) to equal 0, either (x - 1) or (x - 7) must equal what?

Kisachek [45]

$(x - 1)(x - 7) = 0$

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

${x}^{2} - 8x = - 7$

$- 7x = \frac{ - 7}{x}$

$x = \frac{ - 7}{ \frac{x}{ - 7} } \implies x = x \implies x \in \mathbb{R}$

x is true for any number in real domain.

8 0

2 years ago

I have the answer I just have this one question. Will all the paint fit into the bucket?

klasskru [66]

Answer:

brandon

Step-by-step explanation:

Initially, there id 4 pints of green paint in the bucket.

As the green paint is made from equal amounts of yellow paint and blue paint, so, the initial amount of yellow and blue paint in the bucket is 2 pint each.

Let x pints of yellow paint to be added to the bucket to make the desired shade.

Total amount of the paint in the bucket, after addition of pints of yellow color become pint, in which the ampunt of yellow color is pints.

As, the ampunt of yellow color is 80% of the total amount of the mixture, so

So, 6 pints of yellow paint to be added to the bucket to make the desired shade.

Hence, Brandon is correct.

8 0

2 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

2 years ago

What is (0.3)⋅(0.5)?

Harman [31]

Answer:

0.15

Step-by-step explanation:

6 0

2 years ago